What is the advantage of studying Matrices from RD solutions?

RD Sharma Class 12 Solutions Algebra Of Matrices Ex 4.3 is designed by experts to prepare students for their exams. They are simple, easy to understand and cover all concepts from the textbook.

A Matrix is a rectangular array of elements that are arranged in rows and columns. Each element has its own identity, and it need not be related to any other element of the Matrix. A Matrix is represented by M x N. Where N is the number of rows and N is the number of columns. To learn more about Matrices, follow RD Sharma Class 12 Solutions Algebra Of Matrices Ex 4.3.

Name the different types of Matrices

The different types of Matrices are: Square, Symmetric, Diagonal, Identity, Triangular, Orthogonal, etc. These matrices are based on the orientation of their elements. You can refer to Class 12 RD Sharma Chapter 4 Exercise 4.3 Solution to learn more about them.

What is an Idempotent Matrix?

An Idempotent Matrix is a Matrix that gives its value when it is multiplied by itself. For example, if M denotes a Matrix, an Idempotent Matrix can be represented as M2 = M. To learn more about Matrices, you can download RD Sharma Class 12th Exercise 4.3 Solution.

What are the uses of Matrices in real life?

Matrices have a wide range of applications in Engineering and Science. For example, they are used in representing circuits, solving equations, and also in quantum mechanics. Apart from this, they are also used to describe logical data in programming languages. To get a good insight on Matrices, check RD Sharma Class 12 Solutions Chapter 4 Ex 4.3.

RD Sharma Class 12 Exercise 4.3 Algebra of Matrices Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 4.3 Algebra of Matrices Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 4.3 Algebra of Matrices Solutions Maths - Download PDF Free Online

Updated on 20 Jan 2022, 01:49 PM IST

RD Sharma is considered the best book for CBSE Maths students as it contains detailed solutions and is followed by faculties all over the country. It provides a clear explanation for each question, making it easy to RD Sharma Solutions understand and helpful for exam preparations. Students who find it hard can use this book as an excellent way to start building upon their Math skills.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 4 Algebra of Matrices - Other Exercise

Algebra of Matrices Excercise:4.3

Algebra of Matrices exercise 4.3 question 1 (i)

Answer:`

$\begin{bmatrix} a^2 +b^2 &0 \\ 0 & a^2 +b^2 \end{bmatrix}$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: ab-baa-bba$\begin{bmatrix} a &b \\ -b & a \end{bmatrix} \begin{bmatrix} a &-b \\ b &a \end{bmatrix}$
$\begin{aligned} &=\left[\begin{array}{cc} a \times a+b \times b & a \times(-b)+b \times a \\ (-b) \times a+a \times b & (-b) \times(-b)+a \times a \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+b^{2} & -a b+b a \\ -a b+a b & b^{2}+a^{2} \end{array}\right] \end{aligned}$
On simplification we get,
$\begin{bmatrix} a^2+b^2 &0 \\ 0 &b^2+a^2 \end{bmatrix}$

Algebra of Matrices exercise 4.3 question 1 (ii)

Answer:

$\begin{bmatrix} 7 &-2 &5 \\ -7 &10 &3 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix} \begin{bmatrix} 1 &2 &3 \\ -3 &2 &-1 \end{bmatrix}$
$\left[\begin{array}{lll} 1+6 & 2-4 & 3+2 \\ 2-9 & 4+6 & 6-3 \end{array}\right]$
On simplification we get,
$\begin{bmatrix} 7 &-2 &5 \\ -7 &10 &3 \end{bmatrix}$

Algebra of Matrices exercise 4.3 question 1 (iii)

Answer:

$\begin{bmatrix} 14 &0 &42 \\ 18 & -1 & 56\\ 22& -2 &70 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given:$\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]$
$=\left[\begin{array}{lll} 2 \times 1+3 \times 0+4 \times 3 & 2 \times(-3)+3 \times 2+4 \times 0 & 2 \times 5+3 \times 4+4 \times 5 \\ 3 \times 1+4 \times 0+5 \times 3 & 3 \times(-3)+4 \times 2+5 \times 0 & 3 \times 5+4 \times 4+5 \times 5 \\ 4 \times 1+5 \times 0+6 \times 3 & 4 \times(-3)+5 \times 2+6 \times 0 & 4 \times 5+5 \times 4+6 \times 5 \end{array}\right]$
$=\left[\begin{array}{ccc} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{array}\right]$
On simplification we get,
$=\begin{bmatrix} 14 &0 &42 \\ 18 & -1 & 56\\ 22& -2 &70 \end{bmatrix}$

Algebra of Matrices exercise 4.3 question 2 (i)

Answer:

Hence proved $AB\neq BA$
$\begin{bmatrix} 7 &1 \\ 33& 34 \end{bmatrix}\neq \begin{bmatrix} 16 &5 \\ 39& 25 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\begin{bmatrix} 5 &-1 \\ 6 &7 \end{bmatrix}$ and $B=\begin{bmatrix} 2 &1 \\ 3 &4 \end{bmatrix}$
First, we multiply AB matrix
$AB=\begin{bmatrix} 5 &-1 \\ 6 &7 \end{bmatrix}\begin{bmatrix} 2 &1 \\ 3& 4 \end{bmatrix}$
$A B=\left[\begin{array}{cc} 5 \times 2+(-1) \times 3 & 5 \times 1+(-1) \times 4 \\ 6 \times 2+7 \times 3 & 6 \times 1+7 \times 4 \end{array}\right]$
$AB=\begin{bmatrix} 10-3 &5-4 \\ 12+21& 6+28 \end{bmatrix}$
$AB=\begin{bmatrix} 7 &1 \\ 33& 34 \end{bmatrix}$ ...(i)
Again consider
$BA=\begin{bmatrix} 2 &1 \\ 3& 4 \end{bmatrix}\begin{bmatrix} 5 &-1 \\ 6& 7 \end{bmatrix}$
$\begin{aligned} &B A=\left[\begin{array}{ll} 2 \times 5+1 \times 6 & 2 \times(-1)+1 \times 7 \\ 3 \times 5+4 \times 6 & 3 \times(-1)+7 \times 4 \end{array}\right] \\ &B A=\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right] \end{aligned}$
$BA =\begin{bmatrix} 16 &5 \\ 39& 25 \end{bmatrix}$ ...(ii)
From equation (i) and (ii), it is clear that $AB\neq BA$

Algebra of Matrices exercise 4.3 question 2 (ii)

Answer:

Hence proved $AB \neq BA$
$\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \neq\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right]$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A =\begin{bmatrix} -1 &1 &0 \\ 0& -1 &1 \\ 2& 3 &4 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &3 \\ 0& 1 &0 \\ 1& 1 &0 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} -1 &1 &0 \\ 0& -1 &1 \\ 2& 3 &4 \end{bmatrix} \begin{bmatrix} 1 &2 &3 \\ 0& 1 &0 \\ 1& 1 &0 \end{bmatrix}$
$\begin{aligned} &A B=\left[\begin{array}{ccc} (-1) \times 1+1 \times 0+0 \times 1 & -1 \times 2+1 \times 1+0 \times 1 & -1 \times 3+1 \times 0+0 \times 0 \\ 0 \times 1+(-1) \times 0+1 \times 1 & 0 \times 2+(-1) \times 1+1 \times 1 & 0 \times 3+(-1) \times 0+1 \times 0 \\ 2 \times 1+3 \times 0+4 \times 1 & 2 \times 2+3 \times 1+4 \times 1 & 2 \times 3+3 \times 0+4 \times 0 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0 \end{array}\right] \end{aligned}$
$AB =\begin{bmatrix} -1 & -1 &-3 \\ 1& 0 &0 \\ 6 &1 1 &6 \end{bmatrix}$ ...(i)
Now again consider,
$BA=\begin{bmatrix} 1 &2 &3 \\ 0& 1 &0 \\ 1& 1 &0 \end{bmatrix}\begin{bmatrix} -1 &1 &0 \\ 0& -1 &1 \\ 2& 3 &4 \end{bmatrix}$
$\begin{aligned} &B A=\left[\begin{array}{lll} 1 \times(-1)+2 \times 0+3 \times 2 & 1 \times 1+2 \times(-1)+3 \times 3 & 1 \times 0+2 \times 1+3 \times 4 \\ 0 \times(-1)+1 \times 0+0 \times 2 & 0 \times 1+1 \times(-1)+0 \times 3 & 0 \times 0+1 \times 1+0 \times 4 \\ 1 \times(-1)+1 \times 0+0 \times 2 & 1 \times 1+1 \times(-1)+0 \times 3 & 1 \times 0+1 \times 1+0 \times 4 \end{array}\right] \\ B A & =\left[\begin{array}{ccc} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{array}\right] \end{aligned}$
$BA =\begin{bmatrix} 5 &8 &14 \\ 0 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix}$ ...(ii)
From equation (i) and (ii), it is clear that $AB \neq BA$

Algebra of Matrices exercise 4.3 question 2 (iii)

Answer:

Hence proved $AB \neq BA$
$\begin{bmatrix} 3 &1 &0 \\ 1 &1 & 0\\ 1& 4 &0 \end{bmatrix}\neq \begin{bmatrix} 1 &1 &0 \\ 1& 3 & 0\\ 9& 6 &0 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A = \begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix}$ and $B = \begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}$
Consider,
$AB = \begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix}\begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}$
$A B=\left[\begin{array}{lll} 1 \times 0+3 \times 1+0 \times 0 & 1 \times 1+3 \times 0+0 \times 5 & 1 \times 0+3 \times 0+0 \times 1 \\ 1 \times 0+1 \times 1+0 \times 0 & 1 \times 1+1 \times 0+0 \times 5 & 1 \times 0+1 \times 0+0 \times 1 \\ 4 \times 0+1 \times 1+0 \times 0 & 4 \times 1+1 \times 0+0 \times 5 & 4 \times 0+1 \times 0+0 \times 1 \end{array}\right]$
$A B=\left[\begin{array}{lll} 0+3+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 4+0+0 & 0+0+0 \end{array}\right]$
$A B=\begin{bmatrix} 3 &1 &0 \\ 1& 1 & 0\\ 1& 4 &0 \end{bmatrix}$ ...(i)
Now again consider,
$BA = \begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}\begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix}$
$B A=\left[\begin{array}{lll} 0 \times 1+1 \times 1+0 \times 4 & 0 \times 3+1 \times 1+0 \times 1 & 0 \times 0+1 \times 0+0 \times 0 \\ 1 \times 1+0 \times 1+0 \times 4 & 1 \times 3+0 \times 1+0 \times 1 & 1 \times 0+0 \times 0+0 \times 0 \\ 0 \times 1+5 \times 1+1 \times 4 & 0 \times 3+5 \times 1+1 \times 1 & 0 \times 0+5 \times 0+1 \times 0 \end{array}\right]$
$B A=\left[\begin{array}{lll} 0+1+0 & 0+1+0 & 0+0+0 \\ 1+0+0 & 3+0+0 & 0+0+0 \\ 0+5+4 & 0+5+1 & 0+0+0 \end{array}\right]$
$B A=\begin{bmatrix} 1 & 1 &0 \\ 1& 3 & 0\\ 9& 6 &0 \end{bmatrix}$ ...(ii)
From equation (i) and (ii), it is clear that $AB \neq BA$

Algebra of Matrices exercise 4.3 question 3 (i)

Answer:

$AB=\begin{bmatrix} -3 & -4 &1 \\ 8& 13 &9 \end{bmatrix}$ and BA does not exist.
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A=\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &3 \\ 2& 3 &1 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \\ 2& 3 &1 \end{bmatrix}$
$\begin{aligned} &A B=\left[\begin{array}{ccc} 1 \times 1+(-2) \times 2 & 1 \times 2+(-2) \times 3 & 1 \times 3+(-2) \times 1 \\ 2 \times 1+3 \times 2 & 2 \times 2+3 \times 3 & 2 \times 3+3 \times 1 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{array}\right] \end{aligned}$
$AB=\begin{bmatrix} -3 &-4 &1 \\ 8& 13 &9 \end{bmatrix}$
Now consider BA
$BA=\begin{bmatrix} 1 &2 &3 \\ 2& 3 &1 \end{bmatrix}\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix}$
BA doesn’t exist because the number of columns in B is greater than the rows in A.

Algebra of Matrices exercise 4.3 question 3 (ii)

Answer:

$AB= \begin{bmatrix} 12 &17 &22 \\ -4 & -5 & -6\\ -4& -4 &-4 \end{bmatrix}$ and $BA= \begin{bmatrix} 1 &14 \\ -3 &2 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}$ and$B=\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix}$
$\begin{aligned} &A B=\left[\begin{array}{ccc} 3 \times 4+2 \times 0 & 3 \times 5+2 \times 1 & 3 \times 6+2 \times 2 \\ -1 \times 4+0 \times 0 & -1 \times 5+0 \times 1 & -1 \times 6+0 \times 2 \\ -1 \times 4+1 \times 0 & -1 \times 5+1 \times 1 & -1 \times 6+1 \times 2 \end{array}\right] \\ &A B=\left[\begin{array}{lll} 12+0 & 15+2 & 18+4 \\ -4+0 & -5+0 & -6+0 \\ -4+0 & -5+1 & -6+2 \end{array}\right] \end{aligned}$
$AB= \begin{bmatrix} 12 &17 &22 \\ -4 & -5 & -6\\ -4& -4 &-4 \end{bmatrix}$
Again consider
$BA=\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix}\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}$
$\begin{aligned} &B A=\left[\begin{array}{ll} 4 \times 3+5 \times(-1)+6 \times(-1) & 4 \times 2+5 \times 0+6 \times 1 \\ 0 \times 3+1 \times(-1)+2 \times(-1) & 0 \times 2+1 \times 0+2 \times 1 \end{array}\right] \\ B A & =\left[\begin{array}{cc} 12-5-6 & 8+0+6 \\ 0-1-2 & 0+0+2 \end{array}\right] \end{aligned}$
$BA= \begin{bmatrix} 1 &14 \\ -3 &2 \end{bmatrix}$

Algebra of Matrices exercise 4.3 question 3 (iii)

Answer:

AB=11 and $BA =\begin{bmatrix} 0 &0 &0 &0 \\ 1& -1 & 2 &3 \\ 3 &-3 &6 &9 \\ 2 &-2 &4 &6 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A=\begin{bmatrix} 1 &-1 &2 & 3 \end{bmatrix}$ and$B=\begin{bmatrix} 0\\ 1\\ 3\\ 2 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} 1 &-1 &2 & 3 \end{bmatrix}\begin{bmatrix} 0\\ 1\\ 3\\ 2 \end{bmatrix}$
$AB=[ 1 \times 0 +(-1)\times 1 +2\times 3+3\times 2]$
$AB=[ 0-1+6+6]$
$AB=11$
Again consider,
$BA=\begin{bmatrix} 0\\ 1\\ 3\\ 2 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 & 3 \end{bmatrix}$
$B A=\left[\begin{array}{cccc} 0 \times 1 & 0 \times-1 & 0 \times 2 & 0 \times 3 \\ 1 \times 1 & 1 \times(-1) & 1 \times 2 & 1 \times 3 \\ 3 \times 1 & 3 \times(-1) & 3 \times 2 & 3 \times 3 \\ 2 \times 1 & 2 \times(-1) & 2 \times 2 & 2 \times 3 \end{array}\right]$
$BA =\begin{bmatrix} 0 &0 &0 &0 \\ 1& -1 & 2 &3 \\ 3 &-3 &6 &9 \\ 2 &-2 &4 &6 \end{bmatrix}$

Algebra of Matrices exercise 4.3 question 3 (iv)

Answer:

$[a^2+b^2+c^2+d^2+ac+bd]$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given:$\begin{bmatrix} a &b \end{bmatrix}\begin{bmatrix} c\\ d \end{bmatrix}+\begin{bmatrix} a &b& c&d \end{bmatrix}\begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}$
$=[a \times c+b \times d]+[a \times a + b \times b +c \times c + d \times d]$
$=[ac+bd]+[a^2+b^2+c^2+d^2]$
$=[a^2+b^2+c^2+d^2+ac+bd]$

Algebra of Matrices exercise 4.3 question 4 (i)

Answer:

Hence proved $AB \neq BA$
$\begin{bmatrix} 1 &0 &0 \\ 3& -5 &3 \\ 0&0 &1 \end{bmatrix}\neq \begin{bmatrix} -1 &-9 &0 \\ 0& -5 & 0\\ 0& -27 & 1 \end{bmatrix}$
Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\begin{bmatrix} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}, B =\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}$
$A B=\left[\begin{array}{ccc} 1 \times(-2)+3 \times(-1)+(-1) \times(-6) & 1 \times 3+3 \times 2+(-1) \times 9 & 1 \times(-1)+3 \times(-1)+(-1) \times(-4) \\ 2 \times(-2)+(-1) \times(-1)+(-1) \times(-6) & 2 \times 3+(-1) \times 2+(-1) \times 9 & 2 \times(-1)+(-1) \times(-1)+(-1) \times(-4) \\ 3 \times(-2)+0 \times(-1)+(-1) \times(-6) & 3 \times 3+0 \times 2+(-1) \times 9 & 3 \times(-1)+0 \times(-1)+(-1) \times(-4) \end{array}\right]$
$A B=\left[\begin{array}{lll} -2-3+6 & 3+6-9 & -1-3+4 \\ -4+1+6 & 6-2-9 & -2+1+4 \\ -6+0+6 & 9+0-9 & -3+0+4 \end{array}\right]$
$AB=\begin{bmatrix} 1 &0 &0 \\ 3& -5 &3 \\ 0&0 &1 \end{bmatrix}$ ...(i)
Now again consider,
$BA=\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}\begin{bmatrix} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}$
$B A=\left[\begin{array}{lll} -2 \times 1+3 \times 2+(-1) \times 3 & -2 \times 3+3 \times(-1)+(-1) \times 0 & -2 \times(-1)+3 \times(-1)+(-1) \times(-1) \\ -1 \times 1+2 \times 2+(-1) \times 3 & -1 \times 3+2 \times(-1)+(-1) \times 0 & -1 \times(-1)+2 \times(-1)+(-1) \times(-1) \\ -6 \times 1+9 \times 2+(-4) \times 3 & -6 \times 3+9 \times(-1)+(-4) \times 0 & -6 \times(-1)+9 \times(-1)+(-4) \times(-1) \end{array}\right]$
$B A=\left[\begin{array}{ccc} -2+6-3 & -6-3+0 & 2-3+1 \\ -1+4-3 & -3-2+0 & 1-2+1 \\ -6+18-12 & -18-9+0 & 6-9+4 \end{array}\right]$
$BA= \begin{bmatrix} 1 &-9 &0 \\ 0& -5 &0 \\ 0& -27 &1 \end{bmatrix}$ ...(ii)
From equation (i) and (ii), it is clear that $AB \neq BA$

Algebra of Matrices exercise 4.3 question 4 (ii)

Answer:

Hence proved $AB\neq BA$
$\begin{bmatrix} -3 &1 &14 \\ 4& -2 &-17 \\ -5& 1 & 15 \end{bmatrix}\neq \begin{bmatrix} 6 & -4 &1 \\ -5& 3 & -2\\ -5& 1 &1 \end{bmatrix}$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: =$A=\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix}, B = \begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}$
Consider,
$AB=\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix} \begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}$
$\begin{aligned} &A B=\left[\begin{array}{ccc} 10 \times 1+(-4) \times 3+(-1) \times 1 & 10 \times 2+(-4) \times 4+(-1) \times 3 & 10 \times 2+(-4) \times 1+(-1) \times 2 \\ -11 \times 1+5 \times 3+0 \times 1 & -11 \times 2+5 \times 4+0 \times 3 & -11 \times 2+5 \times 1+0 \times 2 \\ 9 \times 1+(-5) \times 3+1 \times 1 & 9 \times 2+(-5) \times 4+1 \times 3 & 9 \times 2+(-5) \times 1+1 \times 2 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} 10-12-1 & 20-16-3 & 20-4-2 \\ -11+15+0 & -22+20+0 & -22+5+0 \\ 9-15+1 & 18-20+3 & 18-5+2 \end{array}\right] \end{aligned}$
$AB =\begin{bmatrix} -3 &1 &14 \\ 4& -2& -17\\ -5 &1 &15 \end{bmatrix}$ ...(i)
Now again consider,
$BA=\begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix}$
$B A=\left[\begin{array}{lll} 1 \times 10+2 \times(-11)+2 \times 9 & 1 \times(-4)+2 \times 5+2 \times(-5) & 1 \times(-1)+2 \times 0+2 \times 1 \\ 3 \times 10+4 \times(-11)+1 \times 9 & 3 \times(-4)+4 \times 5+1 \times(-5) & 3 \times(-1)+4 \times 0+1 \times 1 \\ 1 \times 10+3 \times(-11)+2 \times 9 & 1 \times(-4)+3 \times 5+2 \times(-5) & 1 \times(-1)+3 \times 0+2 \times 1 \end{array}\right]$
$B A=\left[\begin{array}{ccc} 10-22+18 & -4+10-10 & -1+0+2 \\ 30-44+9 & -12+20-5 & -3+0+1 \\ 10-33+18 & -4+15-10 & -1+0+2 \end{array}\right]$
$B A=\begin{bmatrix} 6 &-4 &1 \\ -5 &3 &-2 \\ -5& 1 & 1 \end{bmatrix}$ ...(ii)
From equation (i) and (ii), it is clear that $AB\neq BA$

Algebra of matrices exercise 4.3 question 5 (i)

Answer:

$\begin{bmatrix} 6 &16 &26 \\ -8&- 18 &-28 \end{bmatrix}$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: 1$\left(\left[\begin{array}{cc} 1 & 3 \\ -1 & -4 \end{array}\right]+\left[\begin{array}{cc} 3 & -2 \\ -1 & 1 \end{array}\right]\right)\left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array}\right]$
Firstly, we have to add first two matrix,
$\left[\begin{array}{cc} 1 & 3 \\ -1 & -4 \end{array}\right]+\left[\begin{array}{cc} 3 & -2 \\ -1 & 1 \end{array}\right] \Rightarrow\left[\begin{array}{cc} 1+3 & 3-2 \\ -1-1 & -4+1 \end{array}\right] \Rightarrow\left[\begin{array}{cc} 4 & 1 \\ -2 & -3 \end{array}\right]$
Then, we multiply two matrices
$\begin{bmatrix} 4 &1 \\ -2 &-3 \end{bmatrix} \begin{bmatrix} 1 &3 &5 \\ 2& 4 &6 \end{bmatrix}$
$\begin{aligned} &=\left[\begin{array}{ccc} 4 \times 1+1 \times 2 & 4 \times 3+1 \times 4 & 4 \times 5+1 \times 6 \\ -2 \times 1+(-3) \times 2 & -2 \times 3+(-3) \times 4 & -2 \times 5+(-3) \times 6 \end{array}\right] \\ &=\left[\begin{array}{ccc} 4+2 & 12+4 & 20+6 \\ -2-6 & -6-12 & -10-18 \end{array}\right] \end{aligned}$
On simplification we get
$=\begin{bmatrix} 6 &16 &26 \\ -8&- 18 &-28 \end{bmatrix}$

Algebra of matrices exercise 4.3 question 5 (ii)

Answer:

82
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\begin{bmatrix} 1 &2 &3 \end{bmatrix} \begin{bmatrix} 1 &0 &2 \\ 2& 0& 1\\ 0& 1 &2 \end{bmatrix}\begin{bmatrix} 2\\ \4\\ 6 \end{bmatrix}$
Firstly, we have to multiply first two given matrices,
$\begin{bmatrix} 1 &2 &3 \end{bmatrix} \begin{bmatrix} 1 &0 &2 \\ 2& 0& 1\\ 0& 1 &2 \end{bmatrix}$
$=\begin{bmatrix} 1 \times 1 + 2 \times 2+ 3 \times 0 & 1 \times 0 + 2 \times 0 + 3 \times 1 &1 \times 2 +2 \times + 3 \times 2 \end{bmatrix}$
$=\begin{bmatrix} 1+4 +0 & 0+0+3 &2 + 2+6 \end{bmatrix}$
$=\begin{bmatrix} 5 & 3 &10 \end{bmatrix}$
Now we multiply the above row matrix with third matrix
$\begin{bmatrix} 5 & 3 &10 \end{bmatrix}\begin{bmatrix} 2\\ 4\\ 6 \end{bmatrix}$
$[5\times 2 +3 \times 4 +10 \times 6]=[10+12+60]=82$

Algebra of matrices exercise 4.3 question 5 (iii)

Answer:

$\begin{bmatrix} 0 &-1 &1 \\ 2& 0 &-2 \\ 5& -2 &-3 \end{bmatrix}$
Hint: A matrix can be multiplied by any other matrix that has the same number of rows as the first has columns.
Given:$\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{array}\right]\left(\left[\begin{array}{lll} 1 & 0 & 2 \\ 2 & 0 & 1 \end{array}\right]-\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 0 & 2 \end{array}\right]\right)$
Firstly, we have to subtract the matrix which is inside brackets,
$\begin{aligned} &=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ccc} 1-0 & 0-1 & 2-2 \\ 2-1 & 0-0 & 1-2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 0 \\ 1 & 0 & -1 \end{array}\right] \end{aligned}$
$\begin{aligned} &=\left[\begin{array}{ccc} 1 \times 1+(-1) \times 1 & 1 \times(-1)+(-1) \times 0 & 1 \times 0+(-1) \times(-1) \\ 0 \times 1+2 \times 1 & 0 \times(-1)+2 \times 0 & 0 \times 0+2 \times(-1) \\ 2 \times 1+3 \times 1 & 2 \times(-1)+3 \times 0 & 2 \times 0+3 \times(-1) \end{array}\right] \\ &=\left[\begin{array}{ccc} 1-1 & -1+0 & 0+1 \\ 0+2 & 0+0 & 0-2 \\ 2+3 & -2+0 & 0-3 \end{array}\right] \end{aligned}$
$=\begin{bmatrix} 0 &-1 &1 \\ 2& 0 &-2 \\ 5& -2 &-3 \end{bmatrix}$

Algebra of matrices exercise 4.3 question 6

Answer:

Hence proved $A^2=B^2=C^2=I_2$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}, B =\begin{bmatrix} 1 &0 \\ 0&-1 \end{bmatrix}$ and $C=\begin{bmatrix} 0 &1 \\ 1 &0 \end{bmatrix}$
To Prove:$A^2=B^2=C^2=I_2$
We know that $A^2=AA$
$A^{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 \times 1+0 \times 0 & 1 \times 0+0 \times 1 \\ 0 \times 1+1 \times 0 & 0 \times 0+1 \times 1 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 0+0 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$ ...(i)
Again we know that,
$B^2=BB$
$B^{2}=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{cc} 1 \times 1+0 \times 0 & 1 \times 0+0 \times(-1) \\ 0 \times 1+(-1) \times 0 & 0 \times 0+(-1) \times(-1) \end{array}\right]$
$B^{2}=\left[\begin{array}{ll} 1+0 & 0-0 \\ 0-0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$ ...(ii)
Now consider,
$C^2=CC$
$C^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 0 \times 0+1 \times 1 & 0 \times 1+1 \times 0 \\ 1 \times 0+0 \times 1 & 1 \times 1+0 \times 0 \end{array}\right]$
$C^{2}=\begin{bmatrix} 0+1 & 0+0\\ 0+0 &1+0 \end{bmatrix}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$ ...(iii)
And $I_2=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

where $I_2$ refers to an identity matrix having order 2x2 ( or matrix with two rows and two columns) and 1’s in the main diagonal.
$I_2=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$ ...(iv)
Now, from equation (i), (ii), (iii) and (iv), it is clear that $A^2=B^2=C^2=I_2$

Algebra of matrices exercise 4.3 question 7

Answer:

$\begin{bmatrix} 4 &-20 \\ 38& -10 \end{bmatrix}$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A= \begin{bmatrix} 2 &-1 \\ 3 &2 \end{bmatrix},B = \begin{bmatrix} 0 &4 \\ -1& 7 \end{bmatrix}$
Consider,
$A^2=AA$
$A^{2}=\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 \times 2+(-1) \times 3 & 2 \times(-1)+(-1) \times 2 \\ 3 \times 2+2 \times 3 & 3 \times(-1)+2 \times 2 \end{array}\right]$
$=\left[\begin{array}{ll} 4-3 & -2-2 \\ 6+6 & -3+4 \end{array}\right]=\left[\begin{array}{cc} 1 & -4 \\ 12 & 1 \end{array}\right]$
Now, we have to find $3A^2-2B+I$
Where$I =\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$ identity matrix
$3 A^{2}-2 B+I=3\left[\begin{array}{cc} 1 & -4 \\ 12 & 1 \end{array}\right]-2\left[\begin{array}{cc} 0 & 4 \\ -1 & 7 \end{array}\right]+\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{cc} 3 & -12 \\ 36 & 3 \end{array}\right]-\left[\begin{array}{cc} 0 & 8 \\ -2 & 14 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{cc} 3-0+1 & -12-8+0 \\ 36+2+0 & 3-14+1 \end{array}\right] \end{aligned}$
$=\begin{bmatrix} 4 &-20 \\ 38& -10 \end{bmatrix}$

Algebra of matrices exercise 4.3 question 8

Hence proved $(A-2I)(A-3I)=0$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A =\begin{bmatrix} 4 &2 \\ -1& 1 \end{bmatrix}$
Prove: $(A-2I)(A-3I)=0$
Consider,
$A-2I$
$I =\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$ identity matrix
$\begin{aligned} &A-2 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right] \\ &=\left[\begin{array}{cc} 4-2 & 2-0 \\ -1-0 & 1-2 \end{array}\right] \end{aligned}$
$=\begin{bmatrix} 2 &2 \\ -1 &-1 \end{bmatrix}$ ...(i)
Now consider,
$\begin{aligned} &A-3 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \\ &=\left[\begin{array}{cc} 4-3 & 2-0 \\ -1-0 & 1-3 \end{array}\right] \end{aligned}$
$=\begin{bmatrix} 1 &2 \\ -1 &-2 \end{bmatrix}$ ...(ii)
Now multiply equation (i) & (ii)
$\begin{aligned} &{\left[\begin{array}{cc} 2 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 2 \times 1+2 \times(-1) & 2 \times 2+2 \times(-2) \\ -1 \times 1+(-1) \times(-1) & -1 \times 2+(-1) \times(-2) \end{array}\right]} \\ &=\left[\begin{array}{cc} 2-2 & 4-4 \\ -1+1 & -2+2 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}$
Hence proved$(A-2I)(A-3I)=0$

Algebra of matrices exercise 4.3 question 3 (ix)

Answer: Hence proved
$A^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] and \ A^{3}=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
Consider,$A^2 = AA$
$A^{2}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 \times 1+1 \times 0 & 1 \times 1+1 \times 1 \\ 0 \times 1+1 \times 0 & 0 \times 1+1 \times 1\end{array}\right] \\\\ \\A^{2}=\left[\begin{array}{ll}1+0 & 1+1 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$
Again consider
$\begin{aligned} A^{3} &=A^{2} A \\\\ A^{3} &=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 \times 1+2 \times 0 & 1 \times 1+2 \times 1 \\ 0 \times 1+1 \times 0 & 0 \times 1+1 \times 1 \end{array}\right] \\\\ A^{3} &=\left[\begin{array}{ll} 1+0 & 1+2 \\ 0+0 & 0+1 \end{array}\right] \\ \\A^{3} &=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \end{aligned}$
Hence, proved

Algebra of matrices exercise 4.3 question 10

Answer: Hence proved $A^{2}=0$
Hint:
matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]$
Prove:$A^{2}=0$
Consider,
$A^{2}=A A\\\\ A^{2}=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]\\\\\\ =\left[\begin{array}{cc}a b \times a b+b^{2} \times\left(-a^{2}\right) & a b \times b^{2}+b^{2} \times(-a b) \\ -a^{2} \times a b+(-a b) \times(-a)^{2} & -a^{2} \times b^{2}+(-a b) \times(-a b)\end{array}\right]\\\\ \\=\left[\begin{array}{cc}a^{2} b^{2}-a^{2} b^{2} & a b^{3}-a b^{3} \\ -a^{3} b+a^{3} b & -a^{2} b^{2}+a^{2} b^{2}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ A^{2}=0$
Hence, proved

Algebra of matrices exercise 4.3 question 11

Answer:

Answer: $A^{2}=\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right]$
Hint:
matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Consider,
$A^{2}=A A\\\\ A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right] \\\\ \\ =\left[\begin{array}{cc}(\cos 2 \theta) \times(\cos 2 \theta)+(\sin 2 \theta) \times(-\sin 2 \theta) & \cos 2 \theta \times \sin 2 \theta+\sin 2 \theta \times \cos 2 \theta \\ (-\sin 2 \theta) \times(\cos 2 \theta)+\cos 2 \theta \times(-\sin 2 \theta) & \sin 2 \theta \times(-\sin 2 \theta)+\cos 2 \theta \times \cos 2 \theta\end{array}\right]\\\\\\ =\left[\begin{array}{cc}\cos ^{2} 2 \theta-\sin ^{2} 2 \theta & \cos 2 \theta \sin 2 \theta+\sin 2 \theta \cos 2 \theta \\ -\cos 2 \theta \sin 2 \theta-\sin 2 \theta \cos 2 \theta & -\sin ^{2} 2 \theta+\cos ^{2} 2 \theta\end{array}\right]$
We know that, $\cos ^{2} \theta-\sin ^{2} \theta=\cos ^{2} 2 \theta$

$A^{2}=\left[\begin{array}{cc} \cos (2 \times 2 \theta) & 2 \sin 2 \theta \cos 2 \theta \\ -2 \sin 2 \theta \cos 2 \theta & \cos (2 \times 2 \theta) \end{array}\right]$
Again, we have
$\begin{array}{l} \\\\ \sin 2 \theta=2 \sin \theta \cos \theta\\\\ A^{2}=\left[\begin{array}{cc} \cos 4 \theta & \sin (2 \times 2 \theta) \\ -\sin (2 \times 2 \theta) & \cos 4 \theta \end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc} \cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta \end{array}\right] \end{array}$

Algebra of matrices exercise 4.3 question 12

Answer:
Hence, prove $A B=B A=0_{3 \times 3}$
Hint:
matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right], B=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$
$\text { Prove: } A B=B A=0_{3 \times 3}$
Consider
,$A B=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right] \\\\ \\ A B=\left[\begin{array}{ccc}2(-1)+(-3) 1+(-5)(-1) & 2(3)+(-3)(-3)+(-5) 3 & 2(5)+(-3)(-5)+(-5)(5) \\ (-1)(-1)+(4)(1)+(5)(-1) & (-1)(3)+4(-3)+5(3) & (-1)(5)+4(-5)+(5)(5) \\ (1)(-1)+(-3)(1)+(-4)(-1) & (1)(3)+(-3)(-3)+(-4)(3) & (1)(5)+(-3)(-5)+(-4)(5)\end{array}\right]$
$A B=\left[\begin{array}{ccc}-2-3+5 & 6+9-15 & 10+15-25 \\ 1+4-5 & -3-12+15 & -5-20+25 \\ -1-3+4 & 3+9-12 & 5+15-20\end{array}\right]$
$A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] . . . . (i)$
Again consider
$B A=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\\\\\\ B A=\left[\begin{array}{ccc}(-1)(2)+(3)(-1)+5(1) & (-1)(-3)+(3)(4)+5(-3) & (-1)(-5)+3(5)+5(-4) \\ (1)(2)+(-3)(-1)+(-5)(1) & (1)(-3)+(-3)(4)+(-5)(-3) & (1)(-5)+(-3)(5)+(-5)(-4) \\ (-1)(2)+(3)(-1)+(5)(1) & (-1)(-3)+(3)(4)+(5)(-3) & (-1)(-5)+(3)(5)+(5)(-4)\end{array}\right] \\\\\\\ B A=\left[\begin{array}{ccc}-2-3+5 & 3+12-15 & 5+15-20 \\ 2+3-5 & -3-12+15 & -5-15+20 \\ -2-3+5 & 3+12-15 & 5+15-20\end{array}\right] \\\\ \\B A=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \quad \ldots(i i)$
From equation (i) & (ii)
$A B=B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]_{3 \times 3}$
Hence, proved

Algebra of matrices exercise 4.3 question 13

Answer:
Hence, prove $A B=B A=0_{3 \times 3}$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right], B=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]$
Consider,
$A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right] \\\\ \\ A B=\left[\begin{array}{ccc}0\left(a^{2}\right)+c(a b)+(-b)(a c) & (0)(a b)+(c)\left(b^{2}\right)+(-b)(b c) & (0)(a c)+c(b c)+(-b)\left(c^{2}\right) \\ (-c)\left(a^{2}\right)+(0)(a b)+a(a c) & (-c)(a b)+(0)\left(b^{2}\right)+a(b c) & (-c)(a c)+0(b c)+a\left(c^{2}\right) \\ b\left(a^{2}\right)+(-a)(a b)+0(a c) & (b)(a b)+(-a)\left(b^{2}\right)+0(b c) & (b)(a c)+(-a)(b c)+o\left(c^{2}\right)\end{array}\right]\) \\\\ \\A B=\left[\begin{array}{ccc}0+a b c-a b c & 0+b^{2} c-b^{2} c & 0+b c^{2}-b c^{2} \\ -a^{2} c+0+a^{2} c & -a b c+0+a b c & -a c^{2}+0+a c^{2} \\ a^{2} b-a^{2} b+0 & a b^{2}-a b^{2}+0 & a b c-a b c+0\end{array}\right]$
$A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] . . . . (i)$
Again consider,

$B A=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}\left(a^{2}\right)(0)+(a b)(-c)+(a c)(b) & a^{2}(c)+a b(0)+a c(-a) & a^{2}(-b)+a b(a)+a c(0) \\ a b(0)+b^{2}(-c)+b c(b) & a b(c)+b^{2}(0)+b c(-a) & a b(-b)+b^{2}(a)+b c(0) \\ a c(0)+b c(-c)+c^{2}(b) & a c(c)+b c(0)+c^{2}(-a) & a c(-b)+b c(a)+c^{2}(0)\end{array}\right] \\\\\\ B A=\left[\begin{array}{ccc}0-a b c+a b c & a^{2} c+0-a^{2} c & -a^{2} b+a^{2} b+0 \\ 0-b^{2} c+b^{2} c & a b c+0-a b c & -a b^{2}+a b^{2}+0 \\ 0-b c^{2}+b c^{2} & a c^{2}+0-a c^{2} & -a b c+a b c+0\end{array}\right] \\\\\\ B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \quad \ldots(i i)$
From equation (i) & (ii)
$A B=B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \\\\ A B=B A=0_{3 \times 3}$
Hence, proved

Algebra of matrices exercise 4.3 question 14

Answer: Hence, proved $A B=A$ and $B A=B$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right], B=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]$
Consider,
$A B=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right] \\\\ A B=\left[\begin{array}{ccc}2(2)+(-3)(-1)+(-5)(1) & 2(-2)+(-3)(3)+(-5)(-2) & 2(-4)+(-3)(4)+(-5)(-3) \\ (-1)(2)+4(-1)+5(1) & (-1)(-2)+4(3)+5(-2) & (-1)(-4)+4(4)+5(-3) \\ (1)(2)+(-3)(-1)+(-4)(1) & 1(-2)+(-3)(3)+(-4)(-2) & (1)(-4)+(-3)(4)+(-4)(-3)\end{array}\right] \\\\\\ A B=\left[\begin{array}{ccc}4+3-5 & -4-9+10 & -8-12+15 \\ -2-4+5 & 2+12-10 & 4+16-15 \\ 2+3-4 & -2-9+8 & -4-12+12\end{array}\right]$
$A B=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]=A$
Therefore AB=A
Again consider,
$B A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}2(2)+(-2)(-1)+(-4)(1) & 2(-3)+(-2)(4)+(-4)(-3) & 2(-5)+(-2)(5)+(-4)(-4) \\ -1(2)+3(-1)+4(1) & (-1)(-3)+3(4)+4(-3) & (-1)(-5)+3(5)+4(-4) \\ 1(2)+(-2)(-1)+(-3)(1) & (1)(-3)+(-2)(4)+(-3)(-3) & (1)(-5)+(-2)(5)+(-3)(-4)\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}4+2-4 & -6-8+12 & -10-10+16 \\ -2-3+4 & 3+12-12 & 5+15-16 \\ 2+2-3 & -3-8+9 & -5-10+12\end{array}\right] \\\\ \\\\B A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]=B \\\\ B A=B$

Algebra of matrices exercise 4.3 question 15

Answer: $\left[\begin{array}{ccc}2 & 9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34\end{array}\right]$
Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right], B=\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]$
Find:$A^2 = B^2$
Consider, $A^2$
$A^2 = AA$
$A^{2}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]$
$A^{2}=\left[\begin{array}{ccc}(-1)(-1)+(1)(3)+(-1)(5) & (-1)(1)+(1)(-3)+(-1)(5) & (-1)(-1)+1(3)+(-1) 5 \\ 3(-1)+(-3)(3)+3(5) & 3(1)+(-3)(-3)+(3)(5) & (3)(-1)+(-3)(3)+3(5) \\ 5(-1)+5(3)+5(5) & 5(1)+(-3)(5)+5(5) & 5(-1)+5(3)+5(5)\end{array}\right] \\\\ \\\ A^{2}=\left[\begin{array}{ccc}1+3-5 & -1-3-5 & 1+3-5 \\ -3-9+15 & 3+9+15 & -3-9+15 \\ -5+15+25 & 5-15+25 & -5+15+25\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{ccc}-1 & 9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35\end{array}\right] .... ( i)$
Now, again consider,$B^{2}$
$B^{2}=B B \\\\ B^{2}=\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right] \\\\\\ B^{2}=\left[\begin{array}{ccc}(0)(0)+(4)(1)+3(-1) & 0(4)+4(-3)+3(4) & 0(3)+4(-3)+3(4) \\ 1(0)+(-3)(1)+(-3)(-1) & 1(4)+(-3)(-3)+(-3)(4) & (1)(3)+(-3)(-3)+(-3)(4) \\ (-1)(0)+4(1)+4(-1) & (-1)(4)+4(-3)+4(4) & (-1)(3)+4(-3)+4(4)\end{array}\right] \\\\ \\ B^{2}=\left[\begin{array}{ccc}0+4-3 & 0-12+12 & 0-12+12 \\ 0-3+3 & 4+9-12 & 3+9-12 \\ 0+4-4 & -4-12+16 & -3-12+16\end{array}\right] \\\\ B^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]-----(ii)$
Now by subtracting equation (ii) from (i) we, get
$\begin{array}{l} A^{2}-B^{2}=\left[\begin{array}{ccc} -1 & 9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35 \end{array}\right]-\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ A^{2}-B^{2}=\left[\begin{array}{ccc} -1-1 & 9-0 & -1-0 \\ 3-0 & 27-1 & 3-0 \\ 35-0 & 15-0 & 35-1 \end{array}\right] \\\\ A^{2}-B^{2}=\left[\begin{array}{ccc} -2 & 9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34 \end{array}\right] \end{array}$

Algebra of matrices exercise 4.3 question 16 (i)

Answer: Hence proved $(A B) C=A(B C)$
Hint: Associating property of multiplication is $(A B) C=A(B C)$
Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Prove: $(A B) C=A(B C)$
Given:$A=\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right], B=\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right], C=\left[\begin{array}{c}1 \\ -1\end{array}\right]$
Consider, LHS
$(A B) C=\left(\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right]\right)\left[\begin{array}{c}1 \\ -1\end{array}\right]$

$\begin{aligned} &=\left[\begin{array}{cc}1 \times 1+2 \times(-1)+0(0) & 1(0)+2(2)+0(3) \\ -1 \times 1+(-1)(0)+1(0) & (-1)(0)+0(2)+1(3)\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\ &=\left[\begin{array}{cc}1-2+0 & 0+4+0 \\ -1+0+0 & 0+0+3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\ &=\left[\begin{array}{cc}-1 & 4 \\ -1 & 3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right] \\\\(A B) C &=\left[\begin{array}{c}(-1)(1)+4(-1) \\ (-1)(1)+3(-1)\end{array}\right]=\left[\begin{array}{l}-1-4 \\ -1-3\end{array}\right] \\(A B) C &=\left[\begin{array}{l}-5 \\ -4\end{array}\right] & \ldots(i) \end{aligned}$
Now consider, RHS
$\begin{array}{l} A(B C)=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left(\left[\begin{array}{cc} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c} 1 \\ -1 \end{array}\right]\right)\\\\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} (1)(1)+0(-1) \\ (-1)(1)+2(-1) \\ (0)(1)+3(-1) \end{array}\right]\\\\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 1+0 \\ -1-2 \\ 0-3 \end{array}\right]\\ \\ =\left[\begin{array}{ccc} 1 & 2 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ -3 \\ -3 \end{array}\right]\\\\ =\left[\begin{array}{c} (1)(1)+2(-3)+(0)(-3) \\ (-1)(1)+(0)(-3)+(1)(-3) \end{array}\right]=\left[\begin{array}{c} 1-6+0 \\ -1+0-3 \end{array}\right]\\ \\ A(B C)=\left[\begin{array}{l} -5 \\ -4 \end{array}\right] \end{array}$
From equation (i) and (ii), it is clear that $(A B) C=A(B C)$

Algebra of matrices exercise 4.3 question 16 (ii)

Answer:
Hence proved $(AB)C = A(BC)$
Hint: Associating property of multiplication is $(AB)C = A(BC)$
Given:
$A=\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right], C=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]$
Consider,
$(A B) C=\left(\left[\begin{array}{ccc} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right]\right)\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\$
$=\left[\begin{array}{ccc} 4(1)+2(0)+3(2) & 4(-1)+2(1)+3(-1) & 4(1)+2(2)+3(1) \\ 1(1)+1(0)+2(2) & 1(-1)+1(1)+2(-1) & 1(1)+1(2)+2(1) \\ 3(1)+0+1(2) & 3(-1)+0(1)+1(-1) & 3(1)+0(2)+1(1) \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]$
$\\\\ =\left[\begin{array}{ccc} 4+0+6 & -4+2-3 & 4+4+3 \\ 1+0+4 & -1+1-2 & 1+2+2 \\ 3+0+2 & -3+0-1 & 3+0+1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\$
$=\left[\begin{array}{ccc} 10 & -5 & 11 \\ 5 & -2 & 5 \\ 5 & -4 & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\$
$= {\left[\begin{array}{ccc} 10(1)+(-5)(3)+11(0) & 10(2)+(-5)(0)+11(0) & 10(-1)+(-5)(1)+11(1) \\ 5(1)+(-2)(3)+5(0) & 5(2)+(-2)(0)+5(0) & 5(-1)+(-2)(1)+5(1) \\ 5(1)+(-4)(3)+4(0) & 5(2)+(-4)(0)+4(0) & 5(-1)+(-4)(1)+4(1) \end{array}\right]} \\\\\\ =\left[\begin{array}{ccc} 10-15+0 & 20+0+0 & -10-5+11 \\ 5-6+0 & 10+0+0 & -5-2+5 \\ 5-12+0 & 10+0+0 & -5-4+4 \end{array}\right]$
$(A B) C=\left[\begin{array}{ccc} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{array}\right]---- (1)$
Now consider RHS
$A(B C) =\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left(\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]\right) \\\\\\=\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1-3+0 & 2+0+0 & -1-1+1 \\ 0+3+0 & 0+0+0 & 0+1+2 \\ 2-3+0 & 4-0+0 & -2-1+1 \end{array}\right] \\ \\\\=\left[\begin{array}{ccc} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} -2 & 2 & -1 \\ 3 & 0 & 3 \\ -1 & 4 & -2 \end{array}\right] \\$
$=\left[\begin{array}{lll} 4(-2)+2(3)+3(-1) & 4(2)+2(0)+3(4) & 4(-1)+2(3)+3(-2) \\ 1(-2)+1(3)+2(-1) & 1(2)+1(0)+2(4) & 1(-1)+1(3)+2(-2) \\ 3(-2)+0(3)+1(-1) & 3(2)+0(0)+1(4) & 3(-1)+0(3)+1(-2) \end{array}\right] \\\\\\ =\left[\begin{array}{lll} -8+6-3 & 8+0+12 & -4+6-6 \\ -2+3-2 & 2+0+8 & -1+3-4 \\ -6+0-1 & 6+0+4 & -3+0-2 \end{array}\right] \\\\\\ A(B C) =\left[\begin{array}{lll} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{array}\right] ... (ii)$
From equation (i) and (ii), it is clear that$(AB)C = A(BC)$

Algebra of matrices exercise 4.3 question 17 (i)

Answer:
Hence, verify the distribution of matrix multiplication over matrix addition $A(B+C)=A B+ A C$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{cc}-1 & 0 \\ 2 & 1\end{array}\right], C=\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$
Consider,
$A(B+C) \\\\ =\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left(\left[\begin{array}{cc}-1 & 0 \\ 2 & 1\end{array}\right]+\left[\begin{array}{ll}0 & 1 \\ 1 & -1\end{array}\right]\right) \\\\\\ = \left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left(\left[\begin{array}{cc}-1+0 & 0+1 \\ 2+1 & 1-1\end{array}\right]\right)=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 3 & 0\end{array}\right] \\\\$
$\begin{bmatrix} -1-3 & 1+0 \\ -0+6& 0+0 \end{bmatrix}$
$A(B+C)=\left[\begin{array}{cc} -4 & 1 \\ 6 & 0 \end{array}\right]$

$A B+A C=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 2 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]\\\\\\=\left[\begin{array}{cc}1(-1)+(-1)(2) & 1(0)+(-1)(1) \\0(-1)+2(2) & 0(0)+2(1)\end{array}\right]+\left[\begin{array}{cc}(1)(0)+(-1)(1) & 1(1)+(-1)(-1) \\ 0(0)+2(1) & 0(1)+2(-1)\end{array}\right] \\\\\\=\left[\begin{array}{ll}-1-2 & 0-1 \\ -0+4 & 0+2\end{array}\right]+\left[\begin{array}{ll}0-1 & 1+1 \\ 0+2 & 0-2\end{array}\right]$
$A B+A C=\left[\begin{array}{cc} -4 & 1 \\ 6 & 0 \end{array}\right]$
From equation i & equation ii
$A(B+C)=A B+ A C$

Algebra of matrices exercise 4.3 question 17 (ii)

Answer
: Hence, verify the distribution of matrix multiplication over matrix addition $A(B+C)=A B +A C$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right], B=\left[\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right], C=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \\$

Consider,
$A(B+C) \\ \quad=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left(\left[\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\right) \\\\\\ =\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left(\left[\begin{array}{ll} 0+1 & 1-1 \\ 1+0 & 1+1 \end{array}\right]\right) \\ =\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array}\right] \\$
$A(B+C)=\left[\begin{array}{cc} 2(1)+(-1)(1) & 2(0)+(-1)(2) \\ 1(1)+1(1) & 1(0)+1(2) \\ -1(1)+2(1) & -1(0)+2(2) \end{array}\right]$
$A(B+C)=\left[\begin{array}{cc} 2-1 & 0-2 \\ 1+1 & 0+2 \\ -1+2 & 0+4 \end{array}\right]\\\\ A(B+C)=\left[\begin{array}{cc} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{array}\right]$
Now consider
$A B+A C=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \\\\ \\ =\left[\begin{array}{cc} 2(0)+(-1)(1) & 2(1)+(-1)(1) \\ 1(0)+1(1) & 1(1)+1(1) \\ -1(0)+2(1) & -1(1)+2(1) \end{array}\right]+\left[\begin{array}{cc} 2(1)+(-1)(0) & 2(-1)+(-1)(1) \\ 1(1)+1(0) & 1(-1)+1(1) \\ -1(1)+2(0) & (-1)(-1)+2(1) \end{array}\right]$
$=\left[\begin{array}{cc} 0-1 & 2-1 \\ 0+1 & 1+1 \\ 0+2 & -1+2 \end{array}\right]+\left[\begin{array}{cc} 2+0 & -2-1 \\ 1+0 & -1+1 \\ -1+0 & 1+2 \end{array}\right] \\\\ A B+A C=\left[\begin{array}{cc} -1 & 1 \\ 1 & 2 \\ 2 & 1 \end{array}\right]+\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ -1 & 3 \end{array}\right]$
$A B+A C=\left[\begin{array}{cc} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{array}\right] .... (ii)$
From equation i & ii
$A(B+C)=A B+A C$

Algebra of matrices exercise 4.3 question 18

Answer:
Hence, verified$A(B-C)=A B-A C$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right], B=\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right], C=\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]$
Consider,
$A(B-C)=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left(\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\right)$
$=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left(\left[\begin{array}{ccc}0-1 & 5-5 & -4-2 \\ -2+1 & 1-1 & 3-0 \\ -1-0 & 0+1 & 2-1\end{array}\right]\right)$
$=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -6 \\ -1 & 0 & 3 \\ -1 & 1 & 1 \end{array}\right] \\\\ \\ =\left[\begin{array}{ccc} 1(-1)+0(-1)+(-2)(-1) & 1(0)+0(0)+(-2)(1) & 1(-6)+0(3)+(-2)(1) \\ 3(-1)+(-1)(-1)+0(-1) & 3(0)+(-1)(0)+0(1) & 3(-6)+(-1)(3)+0(1) \\ -2(-1)+1(-1)+1(-1) & -2(0)+1(0)+1(1) & -2(-6)+1(3)+1(1) \end{array}\right] \\\\\ \\ =\left[\begin{array}{ccc} -1-0+2 & 0+0-2 & -6+0-2 \\ -3+1-0 & 0+0+0 & -18-3+0 \\ 2-1-1 & -0+0+1 & 12+3+1 \end{array}\right] \\\\ \\ A(B-C) =\left[\begin{array}{ccc} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{array}\right] ...(i)$
Now consider,
$A B-A C=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} 0+0+2 & 5+0+0 & -4+0-4 \\ 0+2+0 & 15-1+0 & -12-3+0 \\ 0-2-1 & -10+1+0 & 8+3+2 \end{array}\right]-\left[\begin{array}{ccc} 1+0+0 & 5+0+2 & 2+0-2 \\ 3+1+0 & 15-1+0 & 6+0+0 \\ -2-1+0 & -10+1-1 & -4+0+1 \end{array}\right] \\ \\\\ =\left[\begin{array}{ccc} 2 & 5 & -8 \\ 2 & 14 & -15 \\ -3 & -9 & 13 \end{array}\right]-\left[\begin{array}{ccc} 1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3 \end{array}\right]$
$A B-A C=\left[\begin{array}{ccc} 2-1 & 5-7 & -8-0 \\ 2-4 & 14-14 & -15-6 \\ -3+3 & -9+10 & 13+3 \end{array}\right]\\\\ \\ A B-A C =\left[\begin{array}{ccc} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{array}\right] ... (ii)$
From equation i & ii
$A(B-C) = AB - AC$

Algebra of matrices exercise 4.3 question 19

Answer: $a_{43} = 8 \ \ and \ \ a _{ 22} =0$
Hint:
Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
$a_{43}$ means element from 4th row and 3rd column and $\ \ a _{ 22} =0$ means element from 2nd row and 2nd column.

Given:
$\left[\begin{array}{ccc}0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ -3 & 2 \\ 4 & 3\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right]$
Firstly, we will multiply first two matrices
Then,
$A=\left[\begin{array}{cc}0-3+0 & 0+2+0 \\ 4+0+8 & -2+0+6 \\ 0-9+8 & 0+6+6 \\ 8+0+16 & -4+0+12\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right] \\\\\\ A=\left[\begin{array}{cc}-3 & 2 \\ 12 & 4 \\ -1 & 12 \\ 24 & 8\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right] \\\\\\ A=\left[\begin{array}{ccccc}0+6 & -3-6 & 3+8 & -6-8 & 6+0 \\ 0+12 & 12-12 & -12+16 & 24-16 & -24+0 \\ 0+36 & -1-36 & 1+48 & -2-48 & 2+0 \\ 0+24 & 24-24 & -24+32 & 48-32 & -48+0\end{array}\right] \\\\ \\ A=\left[\begin{array}{ccccc}6 & -9 & 11 & -14 & 6 \\ 12 & 0 & 4 & 8 & -24 \\ 36 & -37 & 49 & -50 & 2 \\ 24 & 0 & 8 & 16 & -48\end{array}\right]$
Then,$a_{43} = 8 \ \ and \ \ a _{ 22} =0$

Algebra of matrices exercise 4.3 question 20

Answer: Hence proved,$A^{3}=p I+q A+r A^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]$ and I is the identity matrix of order 3
So,
$I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Consider, LHS side
$A^{2}=A A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q &r \end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+1+0 \\ 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2}\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\$
Now, $A^3 = A^2 A$
$A^{3}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{ccc} 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2} \\ 0+0+p q+p r^{2} & p r+0+q^{2}+q r^{2} & 0+p+q r+q r+r^{3} \end{array}\right]\\\\\\\ A^{3}=\left[\begin{array}{ccc} p & q & r \\ p r & p+q r & q+r^{2} \\ p q+p r^{2} & p r+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right] ....(i)$
Now consider RHS $p I+q A+r A^{2}$
Then put the values in the equation, we get.
$p I+q A+r A^{2}$
$=p\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]+q\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]+r\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\\\ \\ =\left[\begin{array}{ccc} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{array}\right]+\left[\begin{array}{ccc} 0 & q & 0 \\ 0 & 0 & q \\ q p & q^{2} & q r \end{array}\right]+\left[\begin{array}{ccc} 0 & 0 & r \\ r p & r q & r^{2} \\ p r^{2} & r p+q r^{2} & r q+r^{3} \end{array}\right]\\ \\\\\\ =\left[\begin{array}{ccc} p+0+0 & 0+q+0 & 0+0+r \\ 0+0+r p & p+0+q r & 0+q+r^{2} \\ 0+p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\ \\\\ =\left[\begin{array}{ccc} p & q & r \\ r p & p+q r & q+r^{2} \\ p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\$$-----ii$
From equation i & equation ii
$A^{3}=p I+q A+r A^{2}$
Hence, proved.

Algebra of matrices exercise 4.3 question 21

Answer: Hence proved,
$\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
$w^3 = 1$ when w is a complex cube root of unity.
Given: $A= \left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
w is a complex cube root of unity
Prove: LHS=RHS
$\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Consider the LHS\

$=\left[\begin{array}{ccc} 1+w & w+w^{2} & w^{2}+1 \\ w+w^{2} & w^{2}+1 & 1+w \\ w^{2}+w & 1+w^{2} & w+1 \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]$
We know that $1+w+w^{2}=0$ and $w^{3}=1$
$=\left[\begin{array}{ccc} -w^{2} & -1 & -w \\ -1 & -w & -w^{2} \\ -1 & -w & -w^{2} \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]$
Now by simplifying we get,
$\begin{array}{l} \\\\\\\\\\ =\left[\begin{array}{l} -w^{2}-w-w^{3} \\ -1-w^{2}-w^{4} \\ -1-w^{2}-w^{4} \end{array}\right] \\ =\left[\begin{array}{l} -w\left(w+1+w^{2}\right) \\ -1-w^{2}-w^{3} w \\ -1-w^{2}-w^{3} w \end{array}\right] \end{array}$
Again by substituting $1+w+w^{2}=0$ and $w^3 = 1$ in above matrix we get,
$=\left[\begin{array}{c} -w(0) \\ -1-w^{2}-w \\ -1-w^{2}-w \end{array}\right]=\left[\begin{array}{c} 0 \\ -\left(1+w+w^{2}\right) \\ -\left(1+w+w^{2}\right) \end{array}\right]\\\\ =\left[\begin{array}{c} 0 \\ -(0) \\ -(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]$
Therefore LHS=RHS hence proved.

Algebra of matrices exercise 4.3 question 22

Answer: Hence proved, $A^{2}=A$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:$A=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]$
Consider, $A^{2}=A$
$A^{2}=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right] \\\\\\\ =\left[\begin{array}{ccc}4+3-5 & -6-12+15 & -10-15+20 \\ -2-4+5 & 3+16-15 & 5+20-20 \\ 2+3-4 & -3-12+12 & -5-15+16\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]=A$
Hence, $A^{2}=A$

Algebra of matrices exercise 4.3 question23

Algebra of matrices exercise 4.3question23

Answer: Hence proved, $A^2 = l_3$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]$
Prove:$A^2 = l_3$
As we know, $l_3$ is identity matrix of size 3
$l_3 = =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Consider, $A^{2}=A A$
$A^{2}=\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]\\\\\\\ =\left[\begin{array}{ccc}16-3-12 & -4+0+4 & -16+4+12 \\ 12+0-12 & -3+0+4 & -12+0+12 \\ 12-3-9 & -3+0+3 & -12+4+9\end{array}\right]\\\\\\ =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_{3}$

Algebra of matrices exercise 4.3 question 24 (i) maths

Answer:$x = -2$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\left[\begin{array}{lll} 1 & 1 & x \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=0\\$
Firstly, we multiply first two matrices,
$\Rightarrow\left[\begin{array}{lll} 1+0+2 x & 0+2+x & 2+1+0 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{lll} 2 x+1 & 2+x & 3 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=0\\\\\\\ \Rightarrow[2 x+1+2+x+3]=0 \\\\\ \Rightarrow 3 x+6=0\\\\ \Rightarrow 3 x=-6\\\\ \Rightarrow x=-\frac{6}{3}=-2\\\\ \Rightarrow x=-2$

Algebra of matrices exercise 4.3 question 24 (ii) math

Answer: x=13
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ -2 & 4\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ -9 & x\end{array}\right]$
By multiplication of matrices, we have
$\left[\begin{array}{ll}2(1)+3(-2) & 2(-3)+3(4) \\ 5(1)+7(-2) & 5(-3)+7(4)\end{array}\right]=\left[\begin{array}{ll}-4 & 6 \\ -9 & x\end{array}\right] \\\\ \left[\begin{array}{cc}2-6 & -6+12 \\ 5-14 & -15+28\end{array}\right] =\left[\begin{array}{cc}-4 & 6 \\ -9 & x\end{array}\right] \\\\\\ \left[\begin{array}{cc}-4 & 6 \\ -9 & 13\end{array}\right] =\left[\begin{array}{ll}-4 & 6 \\ -9 & x\end{array}\right]$
Then, x=13

Algebra of matrices exercise 4.3 question 25

Answer:
$x=-1$ or $x=-2$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\left[\begin{array}{lll}x & 4 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0$
Firstly, we multiply first two matrices
$\Rightarrow[2 x+4+0 \quad x+0+2 \quad 2 x+8-4]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0 \\\\ \Rightarrow\left[\begin{array}{lll}2 x+4 & x+2 & 2 x+4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0\\\\ \Rightarrow[(2 x+4) x+4(x+2)+(-1)(2 x+4)]=0$
$\Rightarrow 2 x^{2}+4 x+4 x+8-2 x-4=0 \\\\ \Rightarrow 2 x^{2}+6 x+4=0 \\\\ \Rightarrow 2 x^{2}+2 x+4 x+4=0 \\\\ \Rightarrow 2 x(x+1)+4(x+1)=0 \\\\ \Rightarrow(x+1)(2 x+4)=0 \\ \\$
Either $x+1=0$ or $\quad 2 x+4=0$
$\Rightarrow x=-1 \quad \text { or } \quad 2 x=-4 \Rightarrow x=-2$
Hence, $x=-1 \text { or }-2$

Algebra of matrices exercise 4.3 question 26

Answer: x=2
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $\left[\begin{array}{lll}1 & -1 & x\end{array}\right]\left[\begin{array}{ccc}0 & 1 & -1 \\ 2 & 1 & 3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0$
Firstly, we multiply first two matrices
$\Rightarrow\left[\begin{array}{lll}0-2+x & 1-1+x & -1-3+x\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0 \\\\ \Rightarrow\left[\begin{array}{lll}x-2 & x & x-4\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0\\\\ \Rightarrow[0(x-2)+x(1)+(x-4)(1)]=0 \\\\ \Rightarrow 0+x+x-4=0\\\\ \Rightarrow 2 x-4=0\\\\ \Rightarrow 2 x=4\\\\ \Rightarrow x=\frac{4}{2}=2$
Hence, x=2

Algebra of matrices exercise 4.3 question 27 mat

Answer: Hence prove, $A^{2}-A+2 I=0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] and \ \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Prove: $A^{2}-A+2 I=0$
Consider,
$A^{2}=A A \\\\ A^{2}=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2)\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}9-8 & -6+4 \\ 12-8 & -8+4\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]$
Now, put the value of $A^2$ in the given equation $A^2 - A + 2l$, we get
$=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]-\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]+2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ =\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]-\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]\\\\\\ =\left[\begin{array}{ll}1-3+2 & -2+2+0 \\ 4-4+0 & -4+2+2\end{array}\right]$
$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Hence, $A^{2}-A+2 I=0$

Question:28

Algebra of matrices exercise 4.3 question 28

Answer: $\lambda=-7$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.Given:
$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] and I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\ A^{2}=5 A+\lambda I \\\\ \Rightarrow\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+\lambda\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ \Rightarrow\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{cc}\lambda & 0 \\ 0 & \lambda\end{array}\right] \\\\\\\ \Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5 \\ -5 & 10+\lambda\end{array}\right]$
Since, corresponding entries of equal matrices are equal, so
$\Rightarrow 8=15+\lambda \\ \Rightarrow \lambda=8-15\\ \Rightarrow \lambda=-7$

Algebra of matrices exercise 4.3 question 29 math

Answer: Hence, proved $A^{2}-5 A+7 I_{2}=0$
Hint: $I_2$ is the identity matrix of size 2.
$I_{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\$
Given: $A=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]$
Prove:$A^{2}-5 A+7 I_{2}=0$
Consider:
$A^{2}-5 A+7 I_{2} \\\\ =\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]-5\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\\$
$=\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 \quad 3-10+7 \end{array}\right] \\\\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Hence

Algebra of matrices exercise 4.3 question 30 math

Answer: $(2+A)^{3}-19 A=A^{2}-A+8$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: A is a square matrix $A^2 = A$
Consider: $(2+A)^{3}-19 A$
As we know $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$
Then,$\Rightarrow(2+A)^{3}-19 A\\ =8+A^{3}+3(2)^{2} A+3(2) A^{2}-19 A\\ =8+A^{3}+12 A+6 A^{2}-19 A\\ =A^{3}+6 A^{2}-7 A+8\\ =A^{2} A+6(A)-7(A)+8 \quad\left[\right. As \ \ given \left.A^{2}=A\right]\\ =A A+6 A-7 A+8 \quad\left[\right.As \ \ given \left.A^{2}=A\right]\\ =A^{2}-A+8$
Hence, $(2+A)^{3}-19 A=A^{2}-A+8$

Question:31

Algebra of matrices exercise 4.3 question 31

Answer: Hence, proved $A^{3}-4 A^{2}+A=0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
Prove: $A^{3}-4 A^{2}+A=0$
Consider,$A^{2}=A A$
$A^{2}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right] \\\\ A^{3}=A^{2} A\\\\ A^{2} A=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\\\\ A^{3}=\left[\begin{array}{cc}14+12 & 21+24 \\ 8+7 & 12+14\end{array}\right]=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]$
Now putting value of $A^{3}, A^{2}$ and $A$ in the equation $A^{3}-4 A^{2}+A$we get,
$=\left[\begin{array}{cc}26 & 45 \\ 15 & 26\end{array}\right]-4\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{ll} 26 & 45 \\ 15 & 26 \end{array}\right]-\left[\begin{array}{ll} 28 & 48 \\ 16 & 28 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\\\ =0 \\ \text { So, } A^{3}-4 A^{2}+A=0$

Algebra of matrices exercise 4.3 question 32

Answer: Hence, proved $A^{2}-12 A-I=0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]$
I is an identity matrix. so,$I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
To show that $A^{2}-12 A-I=0$
Now, we will find the matrix for $A^2$, we get
$A^{2}=A A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}25+36 & 15+21 \\ 60+84 & 36+49\end{array}\right]\\\\A^{2}=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right] .... (i)$
Now, we will find the matrix for 12A, we get
$12 A=12\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right] \\\\\ 12 A=\left[\begin{array}{cc}60 & 36 \\ 144 & 84\end{array}\right] \quad \ldots(i i)$
So, substituting corresponding values from equation i & ii in $A^{2}-12 A-I$
we get
$=\left[\begin{array}{cc} 61 & 36 \\ 144 & 85 \end{array}\right]-\left[\begin{array}{cc} 60 & 36 \\ 144 & 84 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\$
$\left[\begin{array}{cc} 61-60-1 & 36-36-0 \\ 144-144-0 & 85-84-1 \end{array}\right]$
$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \\ A^{2}-12 A-I=0$
Hence, matrix A is the root of the given equation.

Algebra of matrices exercise 4.3 question 33

Answer: $A^{2}-5 A-14 I=0$
Hint: I is identity matrix so, $I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
Now, we will find the matrix for $A^2$, we get
$A^{2}=A A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right] ... (i)$
Now, we will find the matrix for 5A, we get
$5 A=5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right] \\\\ 5 A=\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right] ... ( ii)$
So, substitute corresponding values from equation i & ii in eqn $A^{2}-5 A-14 I$
we get
$=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ \\\\ =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\\\\\ =\left[\begin{array}{cc}29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14\end{array}\right] \\\\\\ \left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right] = 0$

Algebra of matrices exercise 4.3 question 34 mat

Answer:
$A^{2}-5 A-7 I=0, A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
Prove:$A^{2}-5 A+7 I=0$
Solution: I is identity matrix so
$7 I=7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$
Now, we will find the matrix for $A^2$, we get
$A^{2}=A A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] ...(i)$
Now, we will find the matrix for 5A, we get
$5 A=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ 5 A=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right] ... (ii)$
So, substituting corresponding values from equation i & ii in
$A^{2}-5 A+7 I$
we get
$\begin{array}{l} =\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \\\\ =\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{array}$
$\therefore A^{2}-5 A+7 I=0$
Hence, proved.
We will find $A^4$
$A^{2}-5 A+7 I=0$
Multiply both sides by $A^2$, we get
$A^{2}\left(A^{2}-5 A+7 I\right)=A^{2}(0) \\\\ A^{4}-5 A^{2} A+7 I A^{2}=0 \\\\ A^{4}=5 A^{2} A-7 I A^{2}\\\\ A^{4}=5 A^{2} A-7 A^{2}$
Now we will substitute the corresponding values we get
$A^{4}=5\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\ \\ A^{4}=5\left[\begin{array}{cc}24-5 & 8+10 \\ -15-3 & -5+6\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\\\\\ A^{4}=5\left[\begin{array}{cc}19 & 18 \\ -18 & 1\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\\ A^{4}=\left[\begin{array}{cc}5 \times 19 & 5 \times 18 \\ -18 \times 5 & 1 \times 5\end{array}\right]-\left[\begin{array}{cc}7 \times 8 & 7 \times 5 \\ -5 \times 7 & 3 \times 7\end{array}\right] \\\\\\ A^{4}=\left[\begin{array}{cc}95 & 90 \\ -90 & 5\end{array}\right]-\left[\begin{array}{cc}56 & 35 \\ -35 & 21\end{array}\right]\\\\\\ A^{4}=\left[\begin{array}{cc}95-56 & 90-35 \\ -90+35 & 5-21\end{array}\right] \\\\\\A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$

Algebra of matrices exercise 4.3 question 35

Answer: k=1
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]$ and $A^{2}=k A-2 I_{2}$
$I_ 2$ is an identity matrix of size 2. So, $2 I_{2}=2\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]$
Also given,
$A^{2}=k A-2 I_{2}$
Now, we will find the matrix for $A^2$, we get
$A^{2}=A A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}9-8 & -6+4 \\ 12-8 & -8+4\end{array}\right] \\\\\ A^{2}=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]----i$
Now, we will find the value for kA, we get
$k A=k\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \\\\ k A=\left[\begin{array}{ll}k \times 3 & k \times(-2) \\ k \times 4 & k \times(-2)\end{array}\right]$
So, substituting corresponding values from equation i & ii in
$A^{2}=k A-2 I_{2}$
we get$\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]=\left[\begin{array}{ll}3 k & -2 k \\ 4 k & -2 k\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] \\\\\ \\ \left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]=\left[\begin{array}{ll}3 k-2 & -2 k-0 \\ 4 k-0 & -2 k-2\end{array}\right]$
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal.Hence,
$\begin{array}{l} 3 k=1+2 \\\\ 3 k=3 \\\\ k=\frac{3}{3}=1 \end{array}$
Therefore, the value of k is 1

Algebra of matrices exercise 4.3 question 36

Answer: k=7
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$ and $A^{2}-8 A+k l=0$
I is an identity matrix. So, $k I=k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$
Now, we have to find $A^2$, we get
$A^{2}=A A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{cc}1+0 & 0+0 \\ -1-7 & 0+49\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right] ...(i)$
Now, we will find the matrix 8A, we get
$8 A=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]=\left[\begin{array}{cc}8 \times 1 & 8 \times 0 \\ 8 \times(-1) & 8 \times 7\end{array}\right] \\\\ 8 \boldsymbol{A}=\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right] ... (ii)$
So, substituting corresponding values from equation i & ii in equation

$A^{2}-8 A+k I=0 \\\\ \Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{cc}k & 0 \\ 0 & k\end{array}\right]=0 \\\\\\ \Rightarrow\left[\begin{array}{cc}1-8+k & 0-0+0 \\ -8+8+0 & 49-56+k\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
And to satisfy the above conditions of equality, the corresponding entries of the matrices should be equal
Hence,
1-8+k-0
k=8-1=7
Therefore, the value of k=7

Algebra of matrices exercise 4.3 question 37

Answer: Hence proved$f(A)=0$
Hint: I is an identity matrix.
Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$and $f(x)=x^{2}-2 x-3$
To show that $f(A)=0$
Substitute x=A in f(x) we get
$f(A)=A^{2}-2 A-3 I---i\\\\ f(A)=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]-2\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\\\\ =\left[\begin{array}{ll}1+4 & 2+2 \\ 2+2 & 4+1\end{array}\right]-\left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] =\left[\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right]-\left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\\\\ =\left[\begin{array}{lll}5-2-3 & 4-4-0 \\ 4-4-0 & 5-2-3\end{array}\right]\\\\ =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ =0\\\\$
So, $f (A) = 0$
Hence, proved

Algebra of matrices exercise 4.3 question 38

Answer: $\lambda=4$ and $\mu=-1$
Hint: Iis an identity matrix. Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and$I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\\\\ A^{2}=\lambda A+\mu I$
So,
$\mu I=\mu\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}\mu & 0 \\ 0 & \mu\end{array}\right]$
Now, we will find the matrix for $A^2$, we get
$A^{2}=A A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right] ....(1)$
Now, we will find the matrix for , $\lambda A$we get
$\lambda A=\lambda\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \\\\ \lambda A=\left[\begin{array}{ll}2 \lambda & 3 \lambda \\ 1 \lambda & 2 \lambda\end{array}\right] ....(ii)$
But given, $A^{2}=\lambda A+\mu I$
So, substituting corresponding values from equation i & ii we get
$\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{ll}2 \lambda & 3 \lambda \\ 1 \lambda & 2 \lambda\end{array}\right]+\left[\begin{array}{ll}\mu & 0 \\ 0 & \mu\end{array}\right] \\\\\\ \left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{ll}2 \lambda+\mu & 3 \lambda+0 \\ 1 \lambda+0 & 2 \lambda+\mu\end{array}\right]$
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal.
Hence,
$\lambda+0=4 \\\\ \lambda=4\\\\ 2 \lambda+\mu=7\\\\ 8+\mu=7\\\\ \mu=7-8=-1$
Therefore, the value of $\lambda = 4$ and $\mu = -1$

Algebra of matrices exercise 4.3 question 39

Answer: $x=1 / 5$
Hint: $I_{3}$ is an identity matrix of size
$3, I_{3}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Given:$\left[\begin{array}{ccc}2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{ccc}-x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x\end{array}\right]$ equal to an identity matrix
So, according to given criteria
$\left[\begin{array}{ccc}2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{ccc}-x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Now, we will multiply the two matrices in LHS we get
$=\left[\begin{array}{ccc}-2 x+0+7 x & 2(14 x)+0+7(-4 x) & 2(7 x)+0+7(-2 x) \\ 0+0+0 & 0+1(1)+0 & 0+0+0 \\ 1(-x)+0+1(x) & 1(14 x)+(-2)(1)+1(-4 x) & 1(7 x)+0+1(-2 x)\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}5 x & 28 x-28 x & 14 x-14 x \\ 0 & 0+1+0 & 0\end{array}\right]\\\\$
$=\left[\begin{array}{ccc}5 x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10 x-2 & 5 x\end{array}\right]$
LHS=RHS (given)
$\left[\begin{array}{ccc}5 x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10 x-2 & 5 x\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal.
So, we get
5x=1
x=1/5
So, the value of x is 1/5

Algebra of matrices exercise 4.3 question 40(i)

Answer: x=5 or -3
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{ll}x & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & -3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0$
First, we multiply first two matrices
$\begin {array}{ll}\Rightarrow\left[\begin{array}{ll}x-2 & 0-3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0 \\\ \Rightarrow\left[\begin{array}{ll}x-2 & -3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0\\\\ \Rightarrow[(x-2) x+5(-3)]=0 \\\\ \end{}$
$x^2 - 2x -15 = 0$ then solve quadratic equation
$\begin {array}{ll} \Rightarrow x^{2}-5 x+3 x-15=0 \\\\ \Rightarrow x(x-5)+3(x-5)=0\\ \\ \Rightarrow(x-5)(x+3)=0 \end {}$
$\begin {array}{ll} \Rightarrow Either \ \ x-5=0 \ \ or \ \ x+3=0 \\\\ \Rightarrow \quad x=5 \quad or \quad x=-3 \\\\So, x=5 \ \ or \ \ -3 \end {}$

Algebra of matrices exercise 4.3 question 40 (ii) math

Answer: x=-1
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$
First we multiply first two matrices,$\begin {array}{ll}\Rightarrow[1(1)+2(2)+(1)(1) \quad 1(2)+2(0)+1(0) \quad 1(0)+2(1)+1(2)]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{lll}1+4+1 & 2+0+0 & 0+2+2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0 \end{}$
$\begin {array}{ll}\\\\\Rightarrow\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0\\\\ \Rightarrow[0+2(2)+4(x)]=0\\\\ \Rightarrow 4+4 x=0\\\\ \Rightarrow 4 x=-4 \end{}$
$\begin {array}{ll}\\\\\Rightarrow x = -\frac{4}{4} = -1 \end{}$
Therefore, x=-1

Algebra of matrices exercise 4.3 question 40 (iii)

Answer: $x=\pm 4 \sqrt{3}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
Firstly, we will multiply first two matrices
$\begin {array}{ll} \Rightarrow\left[\begin{array}{lll}x+0-2 & 0-10+0 & 2 x-5-3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0\\\\ \end{}$
$\begin {array}{ll} \Rightarrow[x(x-2)-40+2 x-8]=0\\\\ \Rightarrow x^{2}-2 x-40+2 x-8=0\\\\ \Rightarrow x^{2}-48=0\\\\ \Rightarrow x^{2}=48\\\\ \Rightarrow x=\pm \sqrt{48}\\\\ \Rightarrow x=\pm 4 \sqrt{3}\\\\ \end{}$

Algebra of matrices exercise 4.3 question 40 (iv)

Answer: x=0 or -23/2
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -3 & 0\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0$

Firstly, we will multiply first two matrices
$\begin {array} {ll} \Rightarrow[2 x(1)+3(-3) \quad 2 x(2)+3(0)]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{ll}2 x-9 & 4 x+0\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \end {}$
$\begin {array} {ll}\Rightarrow\left[\begin{array}{ll}2 x-9 & 4 x\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \Rightarrow[(2 x-9) x+4 x(8)]=0\\\\ \Rightarrow[(2 x-9) x+4 x(8)]=0\\\\ \Rightarrow 2 x^{2}-9 x+32 x=0\\\\ \end {}$
$\begin{array}{l} \Rightarrow 2 x^{2}+23 x=0 \\\\ \Rightarrow x(2 x+23)=0 \\\\ \Rightarrow x=0 \quad \text { or } \quad 2 x+23=0 \\\\\end{array}$
$\begin{array}{l} 2 x=-23 \\\\ x=-\frac{23}{2} \\\\\end{array}$

Algebra of matrices exercise 4.3 question41

Answer:
$A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]$
$I_3$ is identity matrix of size 3.
$I_{3}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \ \ and \ \ \ 3 I_{3}=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
Now we will find the matrix for $A^2$ we get
$\begin {array}{ll}A^{2}=A A=\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right] \\\\\ \\A^{2}=\left[\begin{array}{ccc}1+6+0 & 2-8+0 & 0+10+0 \\ 3-12+0 & 6+16-5 & 0-20+15 \\ 0-3+0 & 0+4-3 & 0-5+9\end{array}\right] \\\\\\A^{2}=\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right] \ \ \ \ ...(i) \end{}$
Now, we will find the matrix for 4A, we get
$\begin {array}{ll} 4 A=4\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right] \\\\\ 4 A=\left[\begin{array}{ccc}4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12\end{array}\right] \ \ \ ...(ii) \end{}$
Substituting corresponding values from equation i & ii in the given equation, we get$A^{2}-4 A+3 I_{3} \\\\ =\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right]-\left[\begin{array}{ccc}4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12\end{array}\right]+\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}7-4+3 & -6-8+0 & 10-0+0 \\ -9-12+0 & 17+16+3 & -5-20+0 \\ -3-0+0 & 1+4+0 & 4-12+3\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]$
Hence,
$A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]$

Algebra of matrices exercise 4.3 question42

Answer:
$f(A)=\left[\begin{array}{ccc}4 & 7 & 2 \\ 12 & 19 & 8 \\ 8 & 12 & 3\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]$ and $f(x) = x^ 2 -2x$
Substitute x=A
Then, $f(A)=A^{2}-2 A$
$\begin {array} {ll} f(A)=\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]-2\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]\\\\\\ f(A)=\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3\end{array}\right]-\left[\begin{array}{ccc}0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6\end{array}\right] \end{}$
$\begin{array}{l} f(A)=\left[\begin{array}{ccc} 0+4+0 & 0+5+4 & 0+0+6 \\ 0+20+0 & 4+25+0 & 8+0+0 \\ 0+8+0 & 0+10+6 & 0+0+9 \end{array}\right]-\left[\begin{array}{ccc} 0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6 \end{array}\right] \\\\ f(A)=\left[\begin{array}{ccc} 4 & 9 & 6 \\ 20 & 29 & 8 \\ 8 & 16 & 9 \end{array}\right]-\left[\begin{array}{ccc} 0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6 \end{array}\right] \\\\ f(A)=\left[\begin{array}{ccc} 4-0 & 9-2 & 6-4 \\ 20-8 & 29-10 & 8-0 \\ 8-0 & 16-4 & 9-6 \end{array}\right] \\\\ f(A)=\left[\begin{array}{ccc} 4 & 7 & 2 \\ 12 & 19 & 8 \\ 8 & 12 & 3 \end{array}\right] \end{array}$

Algebra of matrices exercise 4.3 question 43

Answer:
$f(A)=\left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\ \ \ and \ \ f(x)=x^{3}+4 x^{2}-x$
Substitute x=A
Then, $f(A)=A^{3}+4 A^{2}-A ... ( i)$
Now we will find the matrix $A^ 2$
$\begin {array}{ll}A^{2}=\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{lll}0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9+0 & 4+0+0 \\ 0-2+0 & 1+3+0 & 2+0+0\end{array}\right] \\\\ \\A ^{2}=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right] \end{}$

Now we will find matrix $A^ 3$

$\begin{array} {ll} A^{3}=A^{2} A=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{3}=\left[\begin{array}{ccc}0-10+0 & 4+15+0 & 8+0+0 \\ 0+22+4 & -6-33-4 & -12+0+0 \\ 0+8+2 & -2-12-2 & -4+0+0\end{array}\right] \\\\ \\A^{3}=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right] \end{array}$
Put the value of A, $A^2$, $A^3$ in equation i

$f(A)=A^{3}+4 A^{2}-A\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+4\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+\left[\begin{array}{ccc}16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10+16-0 & 19-20-1 & 8+0-2 \\ 26-24-2 & -43+44+3 & -12+16-0 \\ 10-8-1 & -16+16+1 & -4+8-0\end{array}\right] \\\\\\ = \left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]$

Algebra of matrices exercise 4.3 question 44

Answer: Hence, proved A is a root of the polynomial.
Hint: If $f(x)=0$ then x is a root of the polynomial.
Given:
$A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$ and$f(x)=x^{3}-6 x^{2}+7 x+2$
Substitute x=A
Then
$f(A)=A^{3}-6 A^{2}+7 A+2 I_{3}$
Where $I_3$ is identity matrix of size 3
$I_{3}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
First find $A^2$,
$A^{2}=A A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right] \\\\$
$\begin {array}{ll}=\left[\begin{array}{lll}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right] \\\\ =\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right] \end{}$

Now, let us find $A^3$
$A^{3}=A^{2} A=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39\end{array}\right]\\\\\\ =\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]$
Thus,
$f(A)=A^{3}-6 A^{2}+7 A+2 I_{3} \\\\ =\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]+2\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} 21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2 \end{array}\right] \\\\\\ =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=0$
Thus, A is a root of given polynomial

Algebra of matrices exercise 4.3 question 45

Answer: Hence, proved $A^{2}-4 A-5 I=0$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right] \\\\$

Prove: $A^{2}-4 A-5 I=0$

I is identity matrix

$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\begin {array}{ll}5 I=5\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ \mathrm{LHS}=A^{2}-4 A-5 I \end{}$

$\begin {array}{ll}=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ =\left[\begin{array}{lll}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]\\\\\\\ =\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ =\left[\begin{array}{llll}9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5\end{array}\right] \end{}$

$\begin{array}{l} =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \\\\ =0 \\\\ \text { LHS=RHS } \\\\ \end{array}$

Hence, $A^{2}-4 A-5 I=0$

Algebra of matrices exercise 4.3 question 46

Answer: Hence, prove $A^{2}-7 A+10 I_{3}=0$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$A =\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]$

Prove: $A^{2}-7 A+10 I_{3}=0$

$I_3$ is identity matrix of size 3

$I_{3}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\\\\\ \mathrm{LHS}=A^{2}-7 A+10 I_{3}\\\\ =\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]-7\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]+10\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\$

$\begin {array} {ll} =\left[\begin{array}{ccc} 9+2+0 & 6+8+0 & 0+0+0 \\ 3+4+0 & 2+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+25 \end{array}\right]-\left[\begin{array}{ccc} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{array}\right]+\left[\begin{array}{ccc} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{array}\right]-\left[\begin{array}{ccc} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{array}\right]+\left[\begin{array}{ccc} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 11-21+10 & 14-14+0 & 0-0+0 \\ 7-7+0 & 18-28+10 & 0-0+0 \\ 0-0+0 & 0-0+0 & 25-35+10 \end{array}\right]\\\\\\ =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=0\\ \end{}$

Hence, $A^{2}-7 A+10 I_{3}=0$

Algebra of matrices exercise 4.1 question 47

Answer:
$\left[\begin{array}{ll}x & y \\ z & \mu\end{array}\right]=\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}x & y \\ z & \mu\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$
Firstly, we will multiply both matrices in LHS
$\left[\begin{array}{cc}5 x-7 z & 5 y-7 \mu \\ -2 x+3 z & -2 y+3 \mu\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$
Since, corresponding entries of equal matrices are equal, so
$\begin {array }{ll} 5 x-7 z=-16 \quad \ldots(i) \\\\ -2 x+3 z=7 \quad \ldots(ii) \\\\ 5 y-7 \mu=-6 \quad \ldots(iii)\\\\ -2 y+3 \mu=2 \quad \cdots (iv) \end{}$
First solving equation (i) & ii
Multiply equation i by 2 and equation ii by 5 and then add both equations
$\begin {array}{ll} 10 x-14 z=-32\\\\ -10 x+15 z=35\\\\ z=3 \end{}$
Put the value of z in equation i
$\begin {array}{ll} 5 x-7(3)=-16\\\\ 5 x-21=-16\\\\\ 5 x=-16+21 \\\\ 5 x=5 \\\\ x=1 \end{}$
Solving equation iii & iv
Multiply equation iii by 2 and equation iv by 5 and then add both equation
$\begin{array}{l} 10 y-14 \mu=-12\\\\ -10 y+15 \mu=10\\\\ \mu=-2\\ \end{array}$
$\begin {array}{ll} 5 y-7 \mu=-6\\\\ 5 y-7(-2)=-6\\\\ 5 y+14=-6\\\\ 5 y=-20\\\\ y=-4\\ \end{}$
Therefore,$y=-4, \mu=-2, x=1\ \ \ \text { and } \ \ \ z=3 y=-4\\$

Algebra of matrices exercise 4.3 question 48 (i)

Answer:
$A=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 1\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$
The matrix given on the RHS of the equation is a $2 \times 3$ matrix and the matrix given on the LHS of the equation is $2 \times 2$ . So, matrix A has to be $2 \times 3$ matrix.
Since,
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]_{2 \times 2}\ \ \ \ \ \left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]_{2 \times 3}$
So,A is a matrix of order $2 \times 3$
So, let
$A=\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right]$
$\\\\ \Rightarrow\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right] \\\\ \Rightarrow\left[\begin{array}{lll}a+d & b+e & c+f \\ 0+d & 0+e & 0+f\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$
Since, corresponding entries of equal matrices are equal,
So,
$d=1, c=0, f=1$
and
$\\\\a+d=3 \Rightarrow a+1=3 \Rightarrow a=3-1 \Rightarrow a=2 \\\\ b+e=3 \Rightarrow b+0=3 \Rightarrow b=3$
And
$c+f=5 \Rightarrow c+1=5 \Rightarrow c=4$
Hence,
$\\\\A=\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right] \\\\ A=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 1\end{array}\right]$

Algebra of matrices exercise 4.3 question 48 (ii

Answer:
$A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]_{2 \times 3}=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]_{2 \times 3}$
The matrix given on the RHS of the equation is a ${2 \times 3}$ matrix and the matrix given on the LHS of the equation is ${2 \times 3}$ So, matrix A has to be ${2 \times 2}$ matrix.
Now, let
$A=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$
we have,
$\\\\ \Rightarrow\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right] \\\\ \Rightarrow \left[\begin{array}{ccc}a+4 c & 2 a+5 c & 3 a+6 c \\ b+4 d & 2 b+5 d & 3 b+6 d\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
Equating the corresponding elements of the two matrices, we have
$a+4 c=-7,2 a+5 c=-8,3 a+6 c=-9, b+4 d=2,2 b+5 d=4,3 b+6 d=6$
Now,
$\\\\ \Rightarrow a+4 c=-7 \quad \ldots (i) \\\\ \Rightarrow a=-7-4 c\\\\ 2 a+5 c=-8 \quad \ldots (ii)$
Put the value of a in equation ii
$\begin{array}{l} \Rightarrow 2(-7-4 c)+5 c=-8 \\\\ \Rightarrow-14-8 c+5 c=-8 \\\\ \Rightarrow-3 c=6 \\\\ \Rightarrow c=-2 \end{array}$
Now, put value of a in a=-7-4c
$\\\\ \Rightarrow a=-7-4(-2)=-7+8=1 \\\\ \Rightarrow a=1$
Using (iii) in (iv)
$\\\\ \Rightarrow 4-8 d+5 d=4\\\\ \Rightarrow-3 d=0\\\\ \Rightarrow d=0$
Now use the value of d=0 in b+4d=2
$\therefore b=2-4(0)=2$
Thus, a=1, b=2, c=-2, d=0
Hence, the required matrix A is
$\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]$

Algebra of matrices exercise 4.3 question 48 (iii) math

Answer:
$A=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$
We know that two matrices B and C are eligible for the product BC only when number of columns of B is equal to number or rows of C.
So, from the given definition we can consider that the order of matrix A is $1 \times 3$ i.e. we can assume $\therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]_{3 \times 1}\left[\begin{array}{lll}x_{1} & x_{2} & x_{3}\end{array}\right]_{1 \times 3}=\left[\begin{array}{ccc}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]_{3 \times 3}$
$\\\\ \Rightarrow\left[\begin{array}{lll} 4\left(x_{1}\right) & 4\left(x_{2}\right) & 4\left(x_{3}\right) \\ 1\left(x_{1}\right) & 1\left(x_{2}\right) & 1\left(x_{3}\right) \\ 3\left(x_{1}\right) & 3\left(x_{2}\right) & 3\left(x_{3}\right) \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} 4 x_{1} & 4 x_{2} & 4 x_{3} \\ x_{1} & x_{2} & x_{3} \\ 3 x_{1} & 3 x_{2} & 3 x_{3} \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]$

Equating the corresponding element of the two matrices, we have
$x_{1}=-1, x_{2}=2, x_{3}=1$
So, matrix $A=\left[\begin{array}{lll} -1 & 2 & 1\end{array}]\right.$

Algebra of matrices exercise 4.3 question 48 (iv)

Answer: $A = [ -4]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given :
$\\\\ \left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=A \\\\\\ \Rightarrow A=\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\\\\\ \Rightarrow A=[2(-1)+(1)(-1)+3(0) \quad 2(0)+1(1)+3(1) \quad 2(-1)+1(0)+3(1)]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\$
$\\\\ \Rightarrow A=\left[\begin{array}{lll} -2-1+0 & 0+1+3 \quad-2+0+3 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\\\\\ \Rightarrow A=\left[\begin{array}{ccc} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\\\\\ \Rightarrow A=[(-3)(1)+4(0)+1(-1)]\\\\\Rightarrow A=[-3+0-1] \\\\ \Rightarrow A=[-4]$

Algebra of matrices exercise 4.3 question 48 (v)

Answer:
$A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right] A=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$
The matrix given on the LHS of the equation is a $3 \times 2$ matrix and the one given on the RHS of the equation is $3 \times 3$ matrix. So, A has to be $2 \times 3$ matrixNow, let
$\begin{array}{l} \text { Now, let } A=\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\ \end{array}$
$\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 2(a)+(-1) d & 2(b)+(-1)(e) & 2(c)+(-1) f \\ 1(a)+0(d) & 1(b)+0(e) & 1(c)+0(f) \\ -3(a)+4(d) & -3(b)+4(e) & -3(c)+4(f) \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a+0 & b+0 & c+0 \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \end{array}$
Equating the corresponding elements of the two matrices, we have
$\\\\ a = 1, b = -2 , c = -5 \\\\ then \ \ \, 2a - d = -1$
$\begin{array}{l} \Rightarrow 2(1)-d=-1 \\\\ \Rightarrow 2-d=-1 \\\\ \Rightarrow d=2+1=3 \\\\ \Rightarrow 2 b-e=-8 \\\\ \Rightarrow 2(-2)-e=-8 \\\\ \Rightarrow-4-e=-8 \\\\ \Rightarrow e=-4+8=4 \\\\ \Rightarrow e=4 \end{array}$
$\begin{array}{l} \Rightarrow 2 c-f=-10 \\\\ \Rightarrow 2(-5)-f=-10 \\\\ \Rightarrow-10-f=-10 \\\\ \Rightarrow f=-10+10 \\\\ \Rightarrow f=0 \\ \end{array}$
Hence, $a=1, b=-2, c=-5, d=3, e=4 \text { and } f=0 \\$
Therefore, matrix
$A=\left[\begin{array}{ccc} 1 & -2 & -5 \\ 3 & 4 & 0 \end{array}\right]$

Algebra of matrices exercise 4.3 question 48 (vi)

Answer:
$A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0 \\ -5 & 4\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$
The matrix given on the RHS of the equation is a $3 \times 3$ matrix and the one given on the LHS of the equation is $2 \times 3$ matrix. So, A has to be $3 \times 2$ matrix.
Now, let
$\\\\A=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right] \\\\$
$\\\\\Rightarrow\left[\begin{array}{lll} a(1)+b(4) & a(2)+b(5) & a(3)+b(6) \\ c(1)+d(4) & c(2)+5(d) & c(3)+d(6) \\ e(1)+4(f) & e(2)+5(f) & 3(e)+6(f) \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} a+4 b & 2 a+5 b & 3 a+6 b \\ c+4 d & 2 c+5 d & 3 c+6 d \\ e+4 f & 2 e+5 f & 3 e+6 f \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\$
Equating the corresponding elements of the two matrices, we have
$a + 4b = -7$ ...(i)
$2a +5b = -8$ ...(ii)
Now multiply equation i by 2 and subtract equation ii from i
$\begin{array}{l} 2 a+8 b=-14\\ 2 a+5 b=-8\\ 3 b=-6\\ \Rightarrow b=-2\\ \end{array}$
Put value of b=-2 in equation ii we get
$2a + 5 (-2) = -8$
$\\\\ \Rightarrow 2 a-10=-8\\ \Rightarrow 2 a=-8+10\\ \Rightarrow 2 a=2\\ \Rightarrow a=1\\\$
$\\\\ c+4 d=2 ...(iii) \\\\ \Rightarrow 2 c+5 d=4 ... (iv)$
Multiply equation iii by 2 and subtract equation iv from iii

$\\\\ 2 c+8 d=4\\\\ \underline{2 c+5 d=4}\\\\$
$3 d=0 \Rightarrow d=0$
Now, put d=0 in equation iv, we get
$\\\\ \Rightarrow 2 c+5(0)=4\\\\ \Rightarrow 2 c=4\\\\ \Rightarrow c=2\\\\ \Rightarrow e+4 f=11 \cdots (v)\\\\ \Rightarrow 2 e+5 f=10 \cdots(vi) \\\\$
Multiply equation v by 2 and subtract equation vi from v
$\begin{array}{l} 2 e+8 f=22\\\\ \underline{2 e+5 f=10}\\\\ \Rightarrow 3 f=12\\ \\\Rightarrow f=4\\\\ \text { Put } f=4 \text { in equation }(v i)\\\\ \Rightarrow 2 e+5(4)=10\\\\ \Rightarrow 2 e+20=10\\\\ \Rightarrow 2 e=10-20\\\\ \Rightarrow 2 e=-10\\\\ \Rightarrow e=-5\\ \end{array}$

Thus,
$a=1, b=-2, c=2, d=0, e-5 \text { and } f=4$
Hence,
$A=\left[\begin{array}{cc} 1 & -2 \\ 2 & 0 \\ -5 & 4 \end{array}\right]$

Algebra of matrices exercise 4.3 question 49

Answer:
$A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6 I_{2}$
is identity matrix of order 2
$I_{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Now, let
$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$\\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ll}a+b & -2 a+4 b \\ c+d & -2 c+4 d\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$
Since, corresponding entries of equal matrices are equal, so
$\\\\ \Rightarrow a+b=6 \quad \ldots(i) \\\\ \Rightarrow-2 a+4 b=0 \quad \ldots(i i)\\\\ \Rightarrow c+d=0 \quad \ldots(iii)\\\\ \Rightarrow-2 c+4 d=6 \quad \ldots(iv) \\$
Multiply equation i by 4 and subtract equation ii from i
$\\\\ 4 a+4 b=24\\\\ -2 a+4 b=0\\\\ 6 a=24\\\\ a=\frac{24}{6}\\\\ \Rightarrow a=4$

Put a=4 in equation (i)
$\\\\ \Rightarrow a+b=6\\\\ \Rightarrow 4+b=6\\\\ \Rightarrow b=6-4=2\\\\ \Rightarrow b=2$
Multiply equation iii by 2 and add equation iii and iv
$\\\\ 2 c+2 d=0\\\\ -2 c+4 d=6\\\\ 6 d=6\\\\ \Rightarrow d=1$
Put d=1 in equation iii
$\\\\ \Rightarrow c+d=0\\ \Rightarrow c=-1$
Hence
$A=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]$

Algebra of matrices exercise 4.3 question 50

Answer: k=-4
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right] and \ \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] and \ \ A^{2}+I=k A$
Consider $\\\\A^{2}=A A$
$\\\\ A^{2}=\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\\\\\\ =\left[\begin{array}{cc}(-3)(-3)+2(1) & (-3)(2)+2(-1) \\ 1(-3)+(-1)(1) & 1(2)+(-1)(-1)\end{array}\right]\\\\\\ =\left[\begin{array}{cc}9+2 & -6-2 \\ -3-1 & 2+1\end{array}\right] =\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]$
I is an identity matrix
Consider,
$\\\\ k A=k\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]$
$\\\\ k A=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right] ... (ii)$
Substitute all values in equation $A^{2}+I=k A$, we get
$\\\\ \Rightarrow\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}11+1 & -8+0 \\ -4+0 & 3+1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}12 & -8 \\ -4 & 4\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]$
Since, corresponding entries of equal matrices are equal, So
$\begin{array}{l} \Rightarrow-3 k=12 \\\\ \Rightarrow k=-\frac{12}{3} \\\\ \Rightarrow k=-4 \end{array}$
Hence, value of k is -4

Algebra of matrices exercise 4.3 question 51

Answer: Hence proved,
$(A+B)^{2}=A^{2}+B^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], x^{2}=-1$
To prove:
$(A+B)^{2}=A^{2}+B^{2}$
Consider RHS
$\\\\ A^{2}=A A \\\\ B^{2}=B B$
Then,
$\\\\ \Rightarrow A^{2}+B^{2}=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]+\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0(0)+(-x)(x) & 0(-x)+(-x) 0 \\ x(0)+0(x) & x(-x)+0(0)\end{array}\right]+\left[\begin{array}{ll}0(0)+1(1) & 0(1)+1(0) \\ 1(0)+0(1) & 1(1)+0(0)\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0-x^{2} & 0-0 \\ 0+0 & -x^{2}+0\end{array}\right]+\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2} & 0 \\ 0 & -x^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0+0 \\ 0+0 & -x^{2}+1\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0 \\ 0 & -x^{2}+1\end{array}\right]$
Now put
$\\\\x^{2}=-1$
$\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-(-1)+1 & 0 \\ 0 & -(-1)+1\end{array}\right]=\left[\begin{array}{cc}1+1 & 0 \\ 0 & 1+1\end{array}\right] \\\\\\A^{2}+B^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]$
Now taking LHS side
$\\\\(A+B)^{2}=(A+B)(A+B)\\\\ A+B=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{cc} 0+0 & -x+1 \\ x+1 & 0+0 \end{array}\right]=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0(0)+(-x+1)(x+1) & 0(-x+1)+(-x+1)(0) \\ (x+1)(0)+0(x+1) & (x+1)(-x+1)+0(0) \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0+(-x+1)(x+1) & 0+0 \\ 0+0 & (x+1)(-x+1)+0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} -x^{2}-x+x+1 & 0 \\ 0 & -x^{2}+x-x+1 \end{array}\right]$
$(A+B)^{2}=\left[\begin{array}{cc} -x^{2}+1 & 0 \\ 0 & -x^{2}+1 \end{array}\right]$
Now put
$x^2 = -1$
$\\\\ (A+B)^{2}=\left[\begin{array}{cc} -(-1)+1 & 0 \\ 0 & -(-1)+1 \end{array}\right]=\left[\begin{array}{cc} 1+1 & 0 \\ 0 & 1+1 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\\\\\$
Hence proved, LHS=RHS
$(A+B)^{2}=A^{2}+B^{2}$

Algebra of matrices exercise 4.3 question 52

Answer: Hence proved,
$A^{2}+A=A(A+I)$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
To prove:
$A^{2}+A=A(A+I)$
I is identity matrix
Taking LHS
$\\\\ A^{2}+A=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1+0-0 & 0+0-3 & -3+0-3 \\ 2+2+0 & 0+1+3 & -6+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right] \\\\\\=\left[\begin{array}{ccc}1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}1+1 & -3+0 & -6-3 \\ 4+2 & 4+1 & 0+3 \\ 2+0 & 2+1 & 4+1\end{array}\right]=\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right]$
Now, taking RHS
$\\\\ A(A+I)=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left(\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\right)\\\\\\ =\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1+1 & 0+0 & -3+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1\end{array}\right] =\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}2+0+0 & 0+0-3 & -3+0-6 \\ 4+2+0 & 0+2+3 & -6+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right]\\\\ =\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right]$
Therefore, LHS=RHS
Hence proved,$A^{2}+A=A(A+I)$

Algebra of matrices exercise 4.3 question 53

Answer:
$A^{2}-5 A-14 I=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]=0\ \ and \ \ A^{3}=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
First, we solve this $A^{2}-5 A-14 I$
Where I is an identity matrix
Then,
$\\\\ \left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\\\\ \\=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]$
$\\\\ =\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} 29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14 \end{array}\right]\\\\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\\\\\\ A^{2}-5 A-14 I=0\\\\$
Now,
$\\\\ A^{2}-5 A-14 I=0\\\\ A^{2}=5 A+14 I\\\\ A^{3}=A^{2} A=(5 A+14 I) A\\\\ A^{3}=5 A^{2}+14 A\\\\ A^{3}=A^{2} A=5 A^{2}+14 A$
As we know AI=A
$\\\\ A^{3}=5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145+42 & -125-70 \\ -100-56 & 120+28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right]$

Algebra of matrices exercise 4.3 question 54 (i)

Hence proved
$P(x) P(y)=P(x+y)=P(y) P(x)$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$P(x)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$
We have,
$\\\\ P(x) P(y)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right] \\\\\\ P(x) P(y)=\left[\begin{array}{cc}\cos x \cos y-\sin x \sin y & \sin y \cos x+\sin x \cos y \\ -\sin x \cos y-\cos x \sin y & -\sin x \sin y+\cos x \cos y\end{array}\right]$
$P(x) P(y)=\left[\begin{array}{cc} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]=P(x+y)$
Since
$[\cos x \cos y-\sin x \sin y=\cos (x+y), \sin y \cos x+\sin x \cos y \\\\ =\sin (x+y),-\sin x \cos y- \cos x \sin y=-\sin (x+y)]$
We have,
$\\\\ P(y) P(x)=\left[\begin{array}{cc} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]=P(x+y)\\\\ P(x) P(y)=P(x+y)=P(y) P(x)$

Algebra of matrices exercise 4.3 question 54 (ii)

Answer: Hence proved
$P Q=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]=Q P$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$P=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] and \ \ Q=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$
We have,
$\\\\ P Q=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]\\\\ P Q=\left[\begin{array}{ccc}x \times a & 0 & 0 \\ 0 & y \times b & 0 \\ 0 & 0 & z \times c\end{array}\right] \\\\ =\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]$
And we have,
$\begin{array}{l} Q P=\left[\begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]\left[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right] \\\\ =\left[\begin{array}{ccc} a \times x & 0 & 0 \\ 0 & b \times y & 0 \\ 0 & 0 & c \times z \end{array}\right] \end{array}$
$=\left[\begin{array}{ccc} a x & 0 & 0 \\ 0 & b y & 0 \\ 0 & 0 & c z \end{array}\right]$
As xa=ax, yb=by, zc=cz
Therefore,
$P Q=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]=Q P$
Hence proved.

Algebra of matrices exercise 4.3 question 55

Answer:
$A^{2}-5 A+4 I=\left[\begin{array}{ccc} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{array}\right] \text { and } X=\left[\begin{array}{ccc} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2 \end{array}\right]$
Hint: I is an identity matrix,
$I_{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { and } I_{3}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Given:
$A=\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]$
We have to find $A^{2}-5 A+4 I$
$\\\\ =\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]-5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]$
$\\\\=\left[\begin{array}{ccc} 5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{array}\right]\\\\\\ A^{2}-5 A+4 I=\left[\begin{array}{ccc} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{array}\right]$
Now given
$A^{2}-5 A+4 I+X=0$

$\\X=-\left(A^{2}-5 A+4 I\right) \\\\\\ X=(-)\left[\begin{array}{ccc} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{array}\right] \\\\\\ X=\left[\begin{array}{ccc} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2 \end{array}\right]$

Algebra of matrices exercise 4.3 question 56 maths

Answer: Hence proved, $A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$ for all possible integers n.
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
We use the principle of mathematical induction to prove. $A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$
Given:
$A^{n}=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]$
Prove:
$A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$ for all possible integers n …(i)
Solution:
Step 1: put n=1 in eqn (i)
$A^{1}=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]$
So, $A^n$ is true for n=1
Step 2 :let,$A^n$ be true for n=k, then
$A^{k}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right] ... (ii)$
Step 3 : we have to show that $A^{k+1}=\left[\begin{array}{cc}1 & k+1 \\ 0 & 1\end{array}\right]$
So,
$\\A^{k+1}=A^{k} \times A=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \\\\\\\ A^{k+1}=\left[\begin{array}{cc}1+0 & 1+k \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}1 & 1+k \\ 0 & 1\end{array}\right]$
This shows that $A^n$ is true for n=k+1 whenever it is true for n=k
Hence, by the principle of mathematical induction $A^n$ is true for all positive integers n.

Algebra of matrices exercise 4.3 question 57 math

Answer: Hence proved, $A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right]$ for all positive integers n.
Hint: We use the principle of mathematical induction.
Given:
$A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]$
Prove:
$A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right]$ for every positive integer n …(i)
Solution:
step 1:
Put n=1 in eqn (i)
$\\A^{1}=\left[\begin{array}{cc}a^{1} & b\left(\frac{a^{1}-1}{a-1}\right) \\ 0 & 1\end{array}\right]\\\\ A=\left[\begin{array}{cc}a & b\left(\frac{a-1}{a-1}\right) \\ 0 & 1\end{array}\right] \quad \therefore\left\{a^{1}=a\right\} \\\\ A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]$
So, $A^n$ is true for n=1
Step 2: let $A^n$ be true for n=k, so
$A^{k}=\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] ... (ii)$
Step 3: we have to show that
$\\A^{k+1}=A^{k} \times A\\\\ =\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] \times\left[\begin{array}{ll} a & b \\ 0 & 1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1}+0 & a^{k} b+b\left(\frac{a^{k}-1}{a-1}\right) \\ 0+0 & 0+1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1} & \frac{a^{k+1} b-a^{k} b+a^{k} b-b}{a-1} \\ 0 & 1 \end{array}\right]\\\\\\ A^{k+1}=\left[\begin{array}{cc} a^{k+1} & \frac{b\left(a^{k+1}-1\right)}{a-1} \\ 0 & 1 \end{array}\right]$
So,
$A^n$ is true for n=k+1 whenever it is true n=k
Hence, by principle of mathematical induction $A^n$ is true for all positive integer n.

Algebra of matrices exercise 4.3 question 58

Answer: Hence proved,
$A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin n \theta \\ i \sin n \theta & \cos n \theta\end{array}\right]$ for all $n \in N$
Hint: We use the principle of mathematical induction.
Given:
$A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right]$
To show that:
$A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin n \theta \\ i \sin n \theta & \cos n \theta\end{array}\right]$ for all $n \in N$ …(i)
Solution:
step 1: Put n=1 in eqn (i)
$A^{1}=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right]$
So,$A^n$ is true for n=1
Let, $A^n$ is true for n=k, so
$A^{k}=\left[\begin{array}{cc}\cos k \theta & i \sin k \theta \\ i \sin k \theta & \cos k \theta\end{array}\right]$ ... (ii)
Now, we have to show that
$A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \theta & i \sin (k+1) \theta \\ i \sin (k+1) \theta & \cos (k+1) \theta\end{array}\right]$
Now,
$=\left[\begin{array}{cc} \cos k \theta & i \sin k \theta \\ i \sin k \theta & \cos k \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc} \cos k \theta \cos \theta+i^{2} \sin k \theta \sin \theta & i \cos k \theta \sin \theta+i \sin k \theta \cos \theta \\ i \sin k \theta \cos \theta+i \cos k \theta \sin \theta & i^{2} \sin k \theta \sin \theta+\cos \theta \cos k \theta \end{array}\right] \\\\$
$=\left[\begin{array}{cc} \cos (k+1) \theta & i \sin (k+1) \theta \\ i \sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]$
So, $A^n$ is true for n=k+1 whenever it is true for n=k
Hence, by principle of mathematical induction, $A^n$ is true for all positive integers n.

Algebra of matrices exercise 4.3 question 59

Answer: Hence proved,
$A^{n}=\left[\begin{array}{cc}\cos n \alpha+\sin n \propto & \sqrt{2} \sin n \propto \\ -\sqrt{2} \sin n \propto & \cos n \alpha-\sin n \alpha\end{array}\right]$ for all $n \in N$
Hint: We use the principle of mathematical induction.
Given:
$A=\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \propto & \cos \alpha-\sin \alpha\end{array}\right]$
Prove:
$A^{n}=\left[\begin{array}{cc}\cos n \alpha+\sin n \propto & \sqrt{2} \sin n \propto \\ -\sqrt{2} \sin n \propto & \cos n \alpha-\sin n \alpha\end{array}\right]$ for all $n \in N$
Solution:
Step 1: Put n=1 in eqn (i)
$A^{1}=\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \alpha & \cos \alpha-\sin \alpha\end{array}\right]$
$A^n$ is true for n=1
Step 2: Let, $A^n$ is true for n=k
So, $A^{k}=\left[\begin{array}{cc}\cos k \propto+\sin k \propto & \sqrt{2} \sin k \propto \\ -\sqrt{2} \sin k \propto & \cos k \propto-\sin k \alpha\end{array}\right]$
Step 3: Now, we have to show that $A^n$ is true for n=k+1
$A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \propto+\sin (k+1) \propto & \sqrt{2} \sin (k+1) \propto \\ -\sqrt{2} \sin (k+1) \propto & \cos (k+1) \propto-\sin (k+1) \propto\end{array}\right]$
Now, $A^{k+1}=A^{k} A$
$\\=\left[\begin{array}{cc} \cos k \alpha+\sin k \propto & \sqrt{2} \sin k \propto \\ -\sqrt{2} \sin k \propto & \cos k \propto-\sin k \propto \end{array}\right]\left[\begin{array}{cc} \cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \propto & \cos \alpha-\sin \alpha \end{array}\right]\\\\\\ =\left[\begin{array}{cc} (\cos k \propto+\sin k \propto)(\cos \alpha+\sin \propto) & (\cos k \propto+\sin k \propto) \sqrt{2} \sin \propto \\ -2 \sin \alpha \sin k \propto & +\sqrt{2} \sin k \propto(\cos \alpha-\sin \alpha) \\ (\cos \alpha+\sin \propto)(-\sqrt{2} \sin k \propto) & -2 \sin k \propto \sin \propto \\ -\sqrt{2} \sin \alpha(\cos k \propto-\sin k \propto) & +(\cos k \propto-\sin k \propto)(\cos \alpha-\sin \alpha) \end{array}\right]\\$
$=\left[\begin{array}{cc} \cos k \propto \cos \alpha+\sin k \propto \cos \alpha+\cos k \propto \sin \propto & \sqrt{2} \cos k \propto \sin \propto+\sqrt{2} \sin \propto \sin k \propto \\ +\sin \alpha \sin k \propto-2 \sin \propto \sin k \propto & +\sqrt{2} \sin k \propto \cos \propto-\sqrt{2} \sin k \propto \sin \propto \\ -\sqrt{2} \cos \propto \sin k \propto-\sqrt{2} \sin \propto \sin k \propto & -2 \sin k \propto \sin \propto+\cos k \propto \cos \alpha-\cos \alpha \sin k \propto \\ -\sqrt{2} \sin \propto \cos k \propto+\sqrt{2} \sin \propto \sin k \propto & -\sin \propto \cos k \propto+\sin \alpha \sin k \propto \end{array}\right]\\$
$=\left[\begin{array}{cc} \cos \alpha \cos k \propto-\sin \alpha \sin k \propto & \sqrt{2}(\sin k \propto \cos \propto+\cos k \propto \sin \alpha) \\ +\sin \alpha \cos k \propto+\sin k \propto \cos \alpha & \cos k \propto \cos \alpha-\sin k \propto \sin \alpha \\ -\sqrt{2}(\sin k \propto \cos \alpha+\cos k \propto \sin \alpha) & \left.\begin{array}{c} (- \sin k \propto \cos \alpha+\sin \propto \cos k \propto \end{array}\right) \end{array}\right]$
$=\left[\begin{array}{cc} \cos (k+1) \propto+\sin (k+1) \propto & \sqrt{2} \sin (k+1) \propto \\ -\sqrt{2} \sin (k+1) \propto & \cos (k+1) \propto-\sin (k+1) \propto \end{array}\right]$
So, $A^n$ is true for n=k+1 whenever it is true for n=k
Hence, by principle of mathematical induction, $A^n$ is true for $n \in N$

Algebra of matrices exercise 4.3 question 60 math

Answer: Hence proved,
$A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right]$ for every integer n.
Hint: We use the principle of mathematical induction.
Given:
$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
Prove:
$A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right]$ for every positive integer n. …(i)
Solution:
step 1: put n=1 in eqn(i)
$A^{\mathbf{1}}=\left[\begin{array}{ccc}1 & 1 & \frac{1(1+1)}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\ A^{\mathbf{1}}=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
So, $A^n$ is true for n=1
Step 2 : let $A^n$ be true for n=k, so
$A^{k}=\left[\begin{array}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]$
Step 3: we will prove $A^n$ that will be true for n=k+1
Now,$\\A^{k+\mathbf{1}}=A^{k} A\) \(=\left[\begin{array}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1+0+0 & 1+k+0 & 1+k+\frac{k(k+1)}{2} \\ 0+0+0 & 0+1+0 & 0+1+k \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1 & k+1 & \frac{(k+1)(k+2)}{2} \\ 0 & 1 & (k+1) \\ 0 & 0 & 1\end{array}\right]$
Hence, $A^n$ is true for n=k+1 wherever it is true for n=k
So, by principle of mathematical induction $A^n$ is true for all positive integers n.

Algebra of matrices exercise 4.3 question 61 math

Answer: Hence proved, $A^{n+1}=B^{n}(B+(n+1) C$ for every integer $n \in N$.
Hint: We use the principle of mathematical induction.
Given: B, C are n rowed square matrix,
$\\A=B+C, B C=C B, C^{2}=0, \\\\$
$A=B+C$
Squaring both sides, we get
$\\A^{2}=(B+C)^{2}\\\\ A^{2}=(B+C)(B+C)\\\\ A^{2}=B \times B+B C+C B+C$ [using distributive property]
$A^{2}=B^{2}+B C+B C+C^{2}$ [using BC=CB] given and put value of $C^{2}$
$\\A^{2}=B^{2}+2 B C+0\\\\ A^{2}=B^{2}+2 B C \quad \ldots(i)\\\\ A^{2}=B(B+2 C)$
Now consider,$P(n)=A^{n+\mathbf{1}}=B^{n}[B+(n+1) C]$
Step 1: to prove P(1) is true, put n=1
$\\A^{\mathbf{1 + 1}}=B^{\mathbf{1}}[B+(1+1) C]\\\\ A^{2}=B[B+2 C]\\\\ A^{2}=B^{2}+2 B C$
From equation i, P(1)is true
Step 2: suppose P(k) is true
$A^{k+1}=B^{k}[B+(k+1) C] ... (ii)$
Step 3 : now we need to show that P(k+1) is true
That is we need to prove that
$A^{k+2}=B^{k+1}[B+(k+2) C]$
Now,
$\begin{aligned} A^{k+2} &=A^{k} A^{2} \\ &=B^{(k-1)}[B+k C] \times[B(B+2 C)] \\ &=B^{k}[B+k C] \times[B+2 C] \\ &=B^{k}\left[B \times B+B \times 2 C+k C \times B+2 k C^{2}\right] \\ &=B^{k}\left[B^{2}+2 B C+k B C+2 k \times 0\right] \quad\left[\text { since } B C=C B, C^{2}=0\right] \\ &=B^{k}\left[B^{2}+B C(2+k)\right] \\ &=B^{k} \times B[B+(k+2) C] \\ &=B^{k+1}[B+(k+2) C] \end{aligned}$
So, P(n) is true for n=k+1 whenever P(n) is true for n=k.
Therefore, by principle of mathematical induction P(n) is true for all natural number.

Algebra of matrices exercise 4.3 question 62

Answer: Hence proved, $A^{n}=\operatorname{diag}\left(a^{n} \quad b^{n} \quad c^{n}\right)$ for all positive integer n.
Hint: We use the principle of mathematical induction.
Given:$A=\operatorname{diag}\left(\begin{array}{lll}a & b & c\end{array}\right)$
Prove: $A^{n}=\operatorname{diag}\left(a^{n} \quad b^{n} \quad c^{n}\right)$ for all positive integer n …(i)
Solution: step 1:
put n=1 in eqn (i)
$A^{1}=\operatorname{diag}\left(a^{1} \quad b^{1} \quad c^{1}\right) \ \ [ x^1 = x ]$
$A=\operatorname{diag}\left(\begin{array}{lll}a & b & c\end{array}\right)$
So, $A^n$ is true for n=1.
Step 2: let $A^n$ be true for n=k, so
$A^{k}=\operatorname{diag}\left(a^{k} \quad b^{k} \quad c^{k}\right)$
Step 3: now, we have to show that

Now,
$\\ \begin{array}{l} A^{k+1}=A^{k} A \\\\ =d i a g\left(a^{k} \quad b^{k} \quad c^{k}\right) d i a g(a \quad b \quad c) \\\\ A^{k+1}=\left[\begin{array}{ccc} a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k} \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} a^{k} \times a+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+b^{k} \times b+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+c^{k} \times c \end{array}\right] \\\\ =\left[\begin{array}{ccc} a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k+1} \end{array}\right] \\\\\\ A^{k+1}=\operatorname{diag}\left(a^{k+1} \quad b^{k+1} \quad c^{k+1}\right) \end{array}$
So, $A^n$ is true for n=k+1 whenever $A^n$ is true for n=k
Hence, by principle of mathematical induction $A^n$ is true for all positive integers.

Algebra of matrices exercise 4.3 question 63

Answer: Hence proved,$\left(A^{T}\right)^{n}=\left(A^{n}\right)^{T}$ for all $n \in N$.
Hint: We use the principle of mathematical induction.
Given: A is a square matrix.
Prove: $\left(A^{T}\right)^{n}=\left(A^{n}\right)^{T}$ for all $n \in N$.
Let $P(n) =$ $\left(A^{T}\right)^{n}=\left(A^{n}\right)^{T}$ for all $n \in N$. …(i)
Step 1: put n=1 in eqn(i)
$\left(A^{T}\right)^{\mathbf{1}}=\left(A^{1}\right)^{T} \quad\left[x^{\mathbf{1}}=x\right]$ …(ii)
$A^{T}=A^{T}$
Thus, P(n) is true for n=1
Assume that P(n) is true for $n \in N$
$P(k)=\left(A^{T}\right)^{k}=\left(A^{k}\right)^{T}$ …(iii)
To prove that P(k+1) is true, we have
$\left(A^{T}\right)^{k+1} =\left(A^{T}\right)^{k}\left(A^{T}\right)^{1}$
$\\\\=\left(A^{k}\right)^{T}\left(A^{1}\right)^{T} \quad \ \ \ \text { [ using (ii) and (iii) } ]\\\\ =\left(A^{k+1}\right)^{T}$
Thus, P(k+1) is true, whenever P(k) is true.
Hence, by principle of mathematical induction P(n) is true for all $n \in N$.

Algebra of matrices exercise 4.3 question 64

Answer: a=5, b=4 and order of XY and YX are not the same and they are not equal but both are square matrices
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: Matrix X has a+b rows and a+2 columns. Matrix y has b+1 rows and a+3 column both the matrices XY and YX exist.
So, order of matrix $X=(a+b)(a+2)$ order of matrix $Y=(b+1)(a+3)$
Multiplication of matrix YX exists, when the number of columns of Y is equal to the number of rows of X.
$\begin{array}{l} \text { So, } Y_{(b+1) \times(a+3)}{X}_{(a+b) \times(a+2)} \text { exists }\\\\ a+3=a+b\\\\ b=3 \quad \ldots (i) \end{array}$
Multiplication of matrix XY exists, when the number of columns of X is equal to the number of rows of Y.
$\begin{array}{l} X_{(a+b) \times(a+2)} Y_{(b+1) \times(a+3)} \\\\ (a+2)=(b+1) \\\\ a-b=-1 \\\\ a-3=-1 \quad \text { [ using (i) ] }\\\\ a=2 ... (ii)\\ \end{array}$
So order of $X=(a+b) \times(a+2)$
$\\ \qquad \begin{aligned} &=(2+3) \times(2+2) \\ &=5 \times 4 \\ &=(3+1) \times(2+3) \\ &=4 \times 5 \end{aligned}$

Order of $Y=(b+1) \times(a+3)$
$\\ \begin{array}{l} \\=(3+1) \times(2+3) \\ =4 \times 5 \\ \end{array}$
Order of $X_{5 \times 4} Y_{4 \times 5}=5 \times 5 \\$
Order of $Y_{4 \times 5} X_{5 \times 4}=4 \times 4$
So, order of XY and YX are not same and they are not equal but both XY and YX are square matrices.

Algebra of matrices exercise 4.3 question 65 (i)

Answer:
$A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{ll}0 & b \\ 0 & 0\end{array}\right]$, such that $AB \neq BA$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Solution:
Let
$A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right]\ \ and \ \ B=\left[\begin{array}{ll}0 & b \\ 0 & 0\end{array}\right]$
$\begin{array}{l} A B=\left[\begin{array}{ll} a & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\\\\ =\left[\begin{array}{cc} 0+0 & a b+0 \\ 0+0 & 0+0 \end{array}\right]\\\\ A B=\left[\begin{array}{cc} 0 & a b \\ 0 & 0 \end{array}\right] ... (i)\\\\ B A=\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} a & 0 \\ 0 & 0 \end{array}\right]\\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]\\\\ B A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \quad \cdots (ii) \ \end{array}$
From equation i & ii
$AB \neq BA$
When $A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{ll}0 & b \\ 0 & 0\end{array}\right]$

Algebra of matrices exercise 4.3 question 65 (ii)

Answer:
$A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & b \\ 0 & 0\end{array}\right]$, such that $\\\\ A B=0 \ \ but \ \ A \neq 0, B . \neq 0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Solution:
Let
$\\A=\left[\begin{array}{ll}0 & a \\ 0 & 0\end{array}\right] \neq 0\\\\ $$ B=\left[\begin{array}{ll} b & 0 \\ 0 & 0 \end{array}\right] \neq 0\\ $$ A B=\left[\begin{array}{ll}0 & a \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}b & 0 \\ 0 & 0\end{array}\right]\\\\ =\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 0+0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Hence, AB=0
Therefore $A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & b \\ 0 & 0\end{array}\right]$, such that $\\\\ A B=0 \ \ but \ \ A \neq 0, B . \neq 0$

Algebra of matrices exercise 4.3 question 65 (iii)

Answer: $B=\left[\begin{array}{cc}0 & b \\ 0 & 0\end{array}\right] , \ \ A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right],$, such that AB=0 but $BA \neq 0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Solution:
$\begin{array}{l} Let \ \ A=\left[\begin{array}{ll} 0 & a \\ 0 & 0 \end{array}\right], B=\left[\begin{array}{ll} b & 0 \\ 0 & 0 \end{array}\right]\\\\ A B=\left[\begin{array}{ll} 0 & a \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} b & 0 \\ 0 & 0 \end{array}\right]\\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]\\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\\\\ A B=0\\\\ \end{array}$
Now consider,
$\begin{array}{l} \\ B A=\left[\begin{array}{ll} b & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & a \\ 0 & 0 \end{array}\right]\\\\ =\left[\begin{array}{cc} 0+0 & a b+0 \\ 0+0 & 0+0 \end{array}\right]\\\\ =\left[\begin{array}{cc} 0 & a b \\ 0 & 0 \end{array}\right]\\\\\\ B A \neq 0 \end{array}$
Hence,
For AB=0 and , we have $BA \neq 0$
$\ \ A=\left[\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & b \\ 0 & 0\end{array}\right]$

Algebra of matrices exercise 4.3 question 65 (iv)

Answer$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$, such that $A B=A C \ \ but \ \ \ B \neq C, A \neq 0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Solution: Let$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] , \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$
Here,
$A \neq 0, B \neq C$
Consider LHS
$\\A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 0 \\ -1 & 0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Now consider RHS
$\\A C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \ \left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0 +0 & 0+0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
LHS=RHS
So, $A B=A C \ \ for \ \ \ B \neq C, A \neq 0$
We have $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$

Algebra of matrices exercise 4.3 question 66

Answer: $(A+B)^{2}=A^{2}+2 A B+B^{2}$ does not hold
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: A and B be square matrices of same order
Solution: A
$\begin{aligned} (A+B)^{2} =(A+B)(A+B) \\ \end{aligned}$
$= A(A+B)+B(A+B)$ [using distributive property]
$\begin{array}{l} =A A+A B+B A+B B \\ =A^{2}+A B+B A+B^{2} \end{array}$
But,
$(A+B)^{2}=A^{2}+2 A B+B^{2}$ is possible only when AB=BA
As we know $(x+y)^{2}=x^{2}+y^{2}+2 x y$
Here, we can’t say that AB=BA
So,
$(A+B)^{2}=A^{2}+2 A B+B^{2}$ does not hold

Algebra of matrices exercise 4.3 question 67 (i)

Answer: In general matrix multiplication is not always commutative $(A B \neq B A)(A+B)^{2} \neq A^{2}+ 2 A B+B^{2}$
Hint: We use the formula $(x+y)^{2}=x^{2}+y^{2}+2 x y$
Given: A and B be square matrices of same order.
$\\(A+B)^{2} =A^{2}+2 A B+B^{2}\\ \\(A+B)^{2} =(A+B)(A+B) \\\\$
$=A(A+B)+B(A+B)$ [using distributive property ]
$\\=A \times A+A B+B A+B \times B \\\\ =A^{2}+A B+B A+B^{2} \\\\$
$(A+B)^{2} \neq A^{2}+2 A B+B^{2}$
Since, in general matrix multiplication it is not always commutative $(AB \neq BA )$
So, $(A+B)^{2} \neq A^{2}+ 2 A B+B^{2}$

Algebra of matrices exercise 4.3 question 67 (ii)

Answer: In general matrix multiplication is not always commutative$(A B \neq B A)$
so, $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
$(x-y)^{2}=x^{2}+y^{2}-2 x y$
Given: A and B be square matrices of same order $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$
$(A-B)^{2}=(A-B)(A-B) \\\\$
$=A(A-B)-B(A-B) \\$[using distributive property]
$\\=A \times A-A B-B A+B \times B \\\\ =A^{2}-A B-B A+B^{2} \\\\ \neq A^{2}-2 A B+B^{2}$
Since, in general matrix multiplication is not always commutative $A B \neq BA$,
So, $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$

Algebra of matrices exercise 4.3 question 67 (iii) math

Answer: $(A+B)(A-B) \neq A^{2}-B^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: A and B be square matrices of same order $(A+B)(A-B) \neq A^{2}-B^{2}$
$(A+B) (A-B)=A(A-B)+B(A-B)$ [using distributive properties]
$\\=A \times A-A B+B A-B \times B \\\\ =A^{2}-A B+B A-B^{2} \\\\ \neq A^{2}-B^{2}$
Since, in general matrix multiplication is not always commutative $AB \neq BA$
So, $(A+B)(A-B) \neq A^{2}-B^{2}$

Algebra of matrices exercise 4.3 question 68

Answer: A and B are two square matrices with $A B \neq B A$ then $(A B)^{2} \neq A^{2} B^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: A and B be square matrices of order $3 \times 3$
Solution: Let
$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \ \ and \ \ B=\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Here
$A B =\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\\=\left[\begin{array}{lll}1 \times 0+0 \times 2+0 \times 0 & 1 \times 1+0 \times 1+0 \times 0 & 1 \times 0+0 \times 0+0 \times 1 \\ 1 \times 0+1 \times 2+0 \times 0 & 1 \times 1+1 \times 1+0 \times 0 & 1 \times 0+1 \times 0+0 \times 1 \\ 0 \times 0+0 \times 2+1 \times 0 & 0 \times 1+0 \times 1+1 \times 0 & 0 \times 0+0 \times 0+1 \times 1\end{array}\right] \\ \\\\=\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 1\end{array}\right]$
And $B A =\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\\=\left[\begin{array}{lll} 0 \times 1+1 \times 1+0 \times 0 & 0 \times 0+1 \times 1+0 \times 0 & 0 \times 0+1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 1+0 \times 0 & 2 \times 0+1 \times 1+0 \times 0 & 2 \times 0+1 \times 0+0 \times 1 \\ 0 \times 1+0 \times 1+1 \times 0 & 0 \times 0+0 \times 1+1 \times 0 & 0 \times 0+0 \times 0+1 \times 1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 1 & 1 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Here, $AB \neq BA$Now,
$\\(A B)^{2}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\\\ =\left[\begin{array}{lll} 0 \times 0+1 \times 2+0 \times 0 & 0 \times 1+1 \times 2+0 \times 0 & 0 \times 0+1 \times 0+0 \times 1 \\ 2 \times 0+2 \times 2+0 \times 0 & 2 \times 1+2 \times 2+0 \times 0 & 2 \times 0+2 \times 0+0 \times 1 \\ 0 \times 0+0 \times 2+1 \times 0 & 0 \times 1+0 \times 2+1 \times 0 & 0 \times 0+0 \times 0+1 \times 1 \end{array}\right] \\\\\\$
$=\left[\begin{array}{lll} 2 & 2 & 0 \\ 4 & 6 & 0 \\ 0 & 0 & 1 \end{array}\right]$

$\\ A^2=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{lll} 1+0+0 & 0+0+0 & 0+0+0 \\ 1+1+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$\begin{array}{l} B^{2}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 0+2+0 & 0+1+0 & 0+0+0 \\ 0+2+0 & 2+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 2 & 1 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ A^{2} B^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ \end{array}$
$\\ A^{2} B^{2}=\left[\begin{array}{lll} 2+0+0 & 1+0+0 & 0+0+0 \\ 4+2+0 & 2+3+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\\\ A^{2} B^{2}=\left[\begin{array}{lll} 2 & 1 & 0 \\ 6 & 5 & 0 \\ 0 & 0 & 1 \end{array}\right]$
We can see that if we have A and B two square matrices with $AB \neq BA$ then $(A B)^{2} \neq A^{2} B^{2}$

Algebra of matrices exercise 4.3 question 69

Answer: Hence proved $(A+B)^{2}=A^{2}+2 A B+B^{2}$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: A and B be square matrices of same order such that AB=BA
To prove: $(A+B)^{2}=A^{2}+2 A B+B^{2}$
Now, solving LHS gives
$(A+B)^{2}=(A+B)(A+B)$ [using distributive of matrix multiplication over addition]
$\begin{array}{l} =A^{2}+A B+B A+B^{2} \\\\ =A^{2}+A B+A B+B^{2} \quad[\text { as } B A=A B] \\\\ =A^{2}+2 A B+B^{2} \end{array}$
Therefore LHS=RHS
Hence, proved

Algebra of matrices exercise 4.3 question (70)

Answer: Hence proved AB=AC, through $B \neq C, A \neq 0$,
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given
$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right], B=\left[\begin{array}{cc}3 & 1 \\ 5 & 2 \\ -2 & 4\end{array}\right] \quad \mathrm{and} \ \ C=\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right]$
Taking LHS side,
$\begin{aligned} A B=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ 5 & 2 \\ -2 & 4\end{array}\right] \\ =\left[\begin{array}{cc}3+5-2 & 1+2+4 \\ 9+15-6 & 3+6+12\end{array}\right] \end{aligned}$
$A B=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right] ... (i)$
Now taking RHS side,
$\begin{aligned} A C &=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right] \\ &=\left[\begin{array}{cc}4-3+5 & 2+5+0 \\ 12-9+15 & 6+15+0\end{array}\right] \\ A C &=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right] ... (ii)\end{aligned}$
From equation i & ii
AB=AC
Hence, proved

Algebra of matrices exercise 4.3 question (71)

Answer: Bill of A=Rs 157.80, bill of B=Rs 167.40 and bill of C=Rs 281.40
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: Three shopkeepers A, B and C.
A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils.
B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils.
C purchase 11 dozen notebooks, 13 dozen pens and 8 dozen pencils.
Cost of notebook=40 paise = Rs 0.40
Cost of pen=Rs 1.25
Cost of pencil=35 paise = Rs 0.35
The number of items purchased by A, B and C are represented in matrix form as,
[As we know 1 dozen=12 quantity]
$\text { Notebooks } \begin{array}{l} \text { Pens } \end{array} \text { Pencils }\\$
$\begin {aligned} & \begin{array}{l} X=\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{ccc} 12 \times 12 & 5 \times 12 & 6 \times 12 \\ 10 \times 12 & 6 \times 12 & 7 \times 12 \\ 11 \times 12 & 13 \times 12 & 8 \times 12 \end{array}\right] \end{array} \end {aligned}$
$X=\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{ccc} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]$
Now, matrix formed by the cost of each item is given by,
$Y = \left[\begin{array}{l} 0.40 \\ 1.25 \\ 0.35 \end{array}\right]$
Individual bill can be calculated by
$\\X Y=\left[\begin{array}{lll} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]\left[\begin{array}{l} 0.40 \\ 1.25 \\ 0.35 \end{array}\right] \\\\ X Y=\left[\begin{array}{l} 157.80 \\ 167.40 \\ 281.40 \end{array}\right]$
So,
Bill of A=Rs 157.80
Bill of B=Rs 167.40
Bill of C=Rs 281.40

Algebra of matrices exercise 4.3 question (72)

Answer: The total amount the store will receive from selling all the items Rs 1597.20
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Selling price of physics books=Rs 8.30 chemistry books=Rs 3.45 and mathematics=Rs 4.50
Matrix representation of stock of various types of book in the store is given by,
$\begin{array}{l} X=\left[\begin{array}{lll} 10 \times 12 & 8 \times 12 & 5 \times 12 \end{array}\right] \\\\ X=\left[\begin{array}{lll} 120 & 96 & 60 \end{array}\right] \end{array}$
Matrix representation of selling price of each book is given
$Y=\left[\begin{array}{l} 8.30 \\ 3.45 \\ 4.50 \end{array}\right]$
So, total amount received by the store from selling all the items is given by
$\\X Y=\left[\begin{array}{lll} 120 & 96 & 60 \end{array}\right]\left[\begin{array}{l} 8.30 \\ 3.45 \\ 4.50 \end{array}\right] \\\\ X Y=[(120)(8.30)+(96)(3.45)+(60)(4.50)] \\\\ X Y=[996+331.20+270] \\\\ X Y=[1597.20] \\\\$
Total received amount=Rs 1597.20

Algebra of matrices exercise 4.3 question 73

Answer: Amount spent on X=Rs 3400 Amount spent on Y=Rs 7200
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$A=\left[\begin{array}{c} 40 \\ 100 \\ 50 \end{array}\right]$
The number of contacts of each type made in two cities X and Y is given in matrix B as
$B=\left[\begin{array}{ccc}1000 & 500 & 5000 \\ 3000 & 1000 & 10000\end{array}\right] \begin{array}{c}X \\Y \end{array}$
Total amount spent by the group on the two cities X and Y can be given by.
$\\B A=\left[\begin{array}{ccc}1000 & 500 & 5000 \\ 3000 & 1000 & 10000\end{array}\right]\left[\begin{array}{c}40 \\ 100 \\ 50\end{array}\right]\\\\ =\left[\begin{array}{c}40000+50000+250000 \\ 120000+100000+500000\end{array}\right]\\\\ =\left[\begin{array}{l}340000 \\ 720000\end{array}\right]$
As we know 100 paise=Rs 1
Hence,
Amount spent on X=Rs 3400
Amount spent on Y=Rs 7200

Algebra of matrices exercise 4.3 question 74 (i)

Answer: Rs 15000 invested in the first bond and Rs 15000 invested in the second bond
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: Total amount to invest in 2 different types of bonds=Rs 30000
First bond pays 5% interest per year
Second bond pays 7% interest per year
Let Rs x can be invested in the first bond then, the sum of money invested in the second bond will be Rs (30000-x)
First bond pays 5% interest per year second bond pays 7% interest per year
$\therefore$ In order to obtain an annual total interest of Rs 1800, we have
$\Rightarrow[x \quad(30000-x)]\left[\begin{array}{c} \frac{5}{100} \\ \frac{7}{100} \end{array}\right]=1800\\$
$\therefore$ simple interest for 1 year
$=\frac{\text { principal } \times \text { rate }}{100}$

$\\ \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=1800\\\\ \Rightarrow 5 x+210000-7 x=180000\\\\ \Rightarrow 210000-2 x=180000\\\\ \Rightarrow 2 x=210000-180000\\\\ \Rightarrow 2 x=30000\\\\ \Rightarrow x=\frac{30000}{2}\\\\ \Rightarrow x=15000$
Thus, in order to obtain an annual total interest of Rs 1800, the trust should invest Rs 15000 in the first bond and the remaining Rs15000 in the second bond

Algebra of matrices exercise 4.3 question 74 (ii)

Answer: Rs 5000 invested in the first bond and Rs 25000 invested in the second bond
Hint:
$S I=\frac{P R T}{100}$ where P is principal, R is rate, T is time.
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: Total amount to invest in 2 different types of bonds=Rs 30000
First bond pays 5% interest per year
Second bond pays 7% interest per year
Let Rs x can be invested in the first bond then, the sum of money invested in the second bond will be $Rs \ \ (30000 -x )$
First bond pays 5% interest per year second bond pays 7% interest per year
$\therefore$In order to obtain an annual total interest of Rs 2000, we have
$\Rightarrow\left[\begin{array}{ll} x & (30000-x) \end{array}\right]\left[\begin{array}{c} \frac{5}{100} \\ \frac{7}{100} \end{array}\right]=2000\\\\$
$\therefore$ simple interest for 1 year
$=\frac{\text { principal } \times \text { rate }}{100}\\$

$\\ \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=2000\\\\ \Rightarrow 5 x+210000-7 x=200000\\\\ \Rightarrow 210000-2 x=200000\\\\ \Rightarrow 2 x=210000-200000\\\\ \Rightarrow 2 x=10000\\\\ \Rightarrow x=\frac{10000}{2}\\\\ \Rightarrow x=5000$
Thus, in order to obtain an annual total interest of Rs 2000, the trust should invest Rs 5000 in the first bond and the remaining Rs25000 in the second bond.

Algebra of matrices exercise 4.3 question 75

Answer: Total cost incurred by the organization for three villages X, Y and Z are: 30000, 23000 and 39000.
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: The number of attempts made in 3 different villages X, Y and Z are
$\begin{array}{c} X \\ Y \\ Z \end{array}\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]$
An organization tried to generate awareness through (i) house calls, (ii) letters and (iii)announcements
Cost for mode per attempt in house calls Rs 50, letters Rs 20 and announcements Rs 40
The cost for each mode per attempt is represented by $3 \times 1$ matrix
$A=\left[\begin{array}{l}50 \\ 20 \\ 40\end{array}\right]$
The number of attempts made in the three villages X, Y and Z are represented by a $3 \times 3$ matrix.
$B=\left[\begin{array}{ccc}400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150\end{array}\right]$
The total cost incurred by the organization for the three villages separately is given by matrix multiplication.$\\B A=\left[\begin{array}{ccc}400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150\end{array}\right]\left[\begin{array}{c}50 \\ 20 \\ 40\end{array}\right]\\\\\\ B A=\left[\begin{array}{c}400 \times 50+300 \times 20+100 \times 40 \\ 300 \times 50+250 \times 20+75 \times 40 \\ 500 \times 50+400 \times 20+150 \times 40\end{array}\right]\\\\\\ =\left[\begin{array}{l}20000+6000+4000 \\ 15000+5000+3000 \\ 25000+8000+6000\end{array}\right]\\\\\\ =\left[\begin{array}{l}30000 \\ 23000 \\ 39000\end{array}\right]$
Therefore, cost incurred by the organization for the three villages X,Y and Z is 30000, 23000 and 39000.

Algebra of matrices exercise 4.3 question 76

Answer: The total requirements of calories and proteins for the two families are $\left[\begin{array}{ll} 24600 & 576 \\ 15800 & 332 \end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: Two families A and B there are 4 men, 6 women and 2 children in family A, 2 men, 2 women and 4 children in family B.
Daily amount of calories is 2400 for men, 1900 for women, and 1800 for children. 45 grams of protein for men and 55 grams for women and 33 grams for children.
Let F be the family matrix and R be the requirements matrix.
Then,
$\\F= \begin{array}{c}A \\ B \end{array} \left[\begin{array}{lll}4 & 6 & 2 \\ 2 & 2 & 4\end{array}\right] \\\\ R=\begin{array}{c}M \\ W \\ C\end{array}\left[\begin{array}{cc}2400 & 45 \\ 1900 & 55 \\ 1800 & 33\end{array}\right]$
The requirement of calories and proteins of each of the two families is given by the product matrix FR, as matrix F has number of columns equal to number of rows of R thus
$\\F R=\left[\begin{array}{lll}4 & 6 & 2 \\ 2 & 2 & 4\end{array}\right]\left[\begin{array}{ll}2400 & 45 \\ 1900 & 55 \\ 1800 & 33\end{array}\right] \\\\\\ F R=\left[\begin{array}{lc}4 \times 2400+6 \times 1900+2 \times 1800 & 4 \times 45+6 \times 55+2 \times 33 \\ 2 \times 2400+2 \times 1900+4 \times 1800 & 2 \times \times \times 45+2 \times 55+4 \times 33\end{array}\right]\\\\\\ F R=\begin{array}{l}A \\ B\end{array}\left[\begin{array}{ll}24600 & 576 \\ 15800 & 332\end{array}\right]$
We can say that a balanced diet having the required amount of calories and proteins must be taken by each of the family members.

Question:77

Algebra of matrices exercise 4.3 question 77

Answer: The total amount spent by the party in the two cities (in Rs)
$=\begin {array} {c} X \\Y \end{array} \left[\begin{array}{c}9900 \\ 21200\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: The cost per contact(in paisa)
$A=\left[\begin{array}{c}140 \\ 200 \\ 150\end{array}\right] \begin{array}{c}\text { telephone } \\ \text { house calls } \\ \text { letters }\end{array}$
The number of contacts of each type made in two cities X and Y is
$B=\left[\begin{array}{ccc}1000 & 500 & 5000 \\ 3000 & 1000 & 10000\end{array}\right] \begin{array}{c}X\\Y\end{array}$
The total amount of money spent by party in each of the cities for the election is given by the matrix:
$\\B A=\left[\begin{array}{ccc} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{array}\right]\left[\begin{array}{c} 140 \\ 200 \\ 150 \end{array}\right]\\\\\\ =\left[\begin{array}{c} 1000 \times 140+500 \times 200+5000 \times 150 \\ 3000 \times 140+1000 \times 200+10000 \times 150 \end{array}\right]\\\\\\ =\begin{array}{c} X \\ Y \end{array}\left[\begin{array}{c} 140000+100000+750000 \\ 420000+200000+1500000 \end{array}\right]\\\\\\ =\begin{array}{c} X \\ Y \end{array}\left[\begin{array}{c} 990000 \\ 2120000 \end{array}\right]$
The total amount of money spent by party in each of the cities for the election in Rs is given by
$\\=\left(\frac{1}{100}\right)_{Y}^{X}\left[\begin{array}{c} 990000 \\ 2120000 \end{array}\right]\\\\\\ =\begin{array}{c} X \\ Y \end{array}\left[\begin{array}{c} 9900 \\ 21200 \end{array}\right]$ p>

One should consider social activities before casting his/her vote for the party.

RD Sharma Class 12th Exercise 4.3 deals with the topic Algebra of Matrices. Matrices is a fun chapter for students as it is relatively simple, and the sums are pretty straightforward compared to other units. In this chapter, you will learn about different types of Matrix and solve certain sums related to them.

You will learn about types of Matrices at the beginning of RD Sharma Class 12th Exercise 4.3. Next, you will solve basic questions pertaining to matrix addition and subtraction.There are approximately 77 - Level 1 questions in this exercise that will help you understand the chapter. This might seem a lot for students, but Career360 has covered them with RD Sharma Class 12 Chapter 4 Exercise 4.3 material.

These are the topics that you will learn in this chapter:

Associative and distributive properties of Matrix Multiplication
Sums to prove LHS = RHS
Roots of an Equation
Algebra to find the value of variables
PMI Application based questions

The first 30 questions have low complexity and contain the basic concepts that you have learned. After solving a few questions, you can refer to the solved material to understand each question as they have the same concept.

The last 17 questions are fundamental in terms of exams. These are theory-based questions that you will have to solve by understanding and implementing the concept. As solving all the RD Sharma Class 12th Exercise 4.3 solutions is impossible, you should consider dividing these questions into parts and solving each part every day.

RD Sharma Class 12th Exercise 4.3, provided by Career360, is an excellent source for students to cover all concepts and save a lot of time. Thousands of students have already started preparing this material. So stop wasting more time and be a part of the team. As this material is free for everyone, you can take advantage of it to be the brightest in your class.

Chapter-wise RD Sharma Class 12 Solutions