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RD Sharma Class 12 Exercise 4.2 Algebra of Matrices Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 4.2 Algebra of Matrices Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 4.2 Algebra of Matrices Solutions Maths - Download PDF Free Online

Updated on 20 Jan 2022, 01:43 PM IST

We are pretty sure you have already heard about RD Sharma solutions from your teachers and friends. RD Sharma class 12th exercise 4.2 especially is an excellent guide for students who are in class 12. With the help of RD Sharma class 12 chapter 4 exercise 4.2, they will be able to improve their knowledge of their subjects and prepare themselves for their exams. These answers prepared by experts from all over the country span the 4th chapter of the book, Algebra Of Matrices. This exercise contains detailed information on addition, subtraction, multiplication of matrices, types of matrices like skew-symmetric matrix, null matrix, and symmetric matrix, application-based question using matrix method, and how to perform Algebra calculations with them. RD Sharma class 12th exercise 4.2 contains 22 questions on two levels. The RD Sharma Solutions have a lot of benefits for students, which we will discuss below:-

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

Algebra of Matrices Excercise: 4.2

Algebra of Matrices Exercise 4.2 Question 1(i)

Answer:$\begin{bmatrix} 1 &2 \\ 2 & 7 \end{bmatrix}$
Hint: Add the two matrices simply.
Given: $\begin{bmatrix} 3 &-2 \\ 1 & 4 \end{bmatrix}+\begin{bmatrix} -2 &4 \\ 1& 3 \end{bmatrix}$
Solution:
$\begin{bmatrix} 3 &-2 \\ 1 & 4 \end{bmatrix}+\begin{bmatrix} -2 &4 \\ 1& 3 \end{bmatrix}$ $= \begin{bmatrix} 3-2 &-2+4 \\ 1+1 & 4+3 \end{bmatrix}$
$= \begin{bmatrix} 1&2 \\ 2 & 7 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 1 (ii)

Answer:$\begin{bmatrix} 3 & -1 & 6\\ 2& 9& 6\\ -1 & -1 &6 \end{bmatrix}$
Hint: Add the two matrices simply.
Given:$\begin{bmatrix} 2 & 1& 3\\ 0& 3& 5\\ -1 & 2 &5 \end{bmatrix}+\begin{bmatrix} 1 &-2 &3 \\ 2 &6 &1 \\ 0 &-3 &1 \end{bmatrix}$
Solution:
$\begin{bmatrix} 2 & 1& 3\\ 0& 3& 5\\ -1 & 2 &5 \end{bmatrix}+\begin{bmatrix} 1 &-2 &3 \\ 2 &6 &1 \\ 0 &-3 &1 \end{bmatrix}$ $= \begin{bmatrix} 2+1 & 1-2& 3+3\\ 0+2& 3+6& 5+1\\ -1+0 & 2-3 &5+1 \end{bmatrix}$
$= \begin{bmatrix} 3 & -1 & 6\\ 2& 9& 6\\ -1 & -1 &6 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 2(i)

Answer: $\begin{bmatrix} 1 &-1 \\ 12 &-11 \end{bmatrix}$
Hint: Multiply the A matrix with 2, then multiply B matrix with 3. Then, subtract it from A.
Given:
$A= \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ , $B= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$ and $C= \begin{bmatrix}-2 &5 \\ 3 &4 \end{bmatrix}$
Here, we have to compute $2A-3B$
Solution:
$2A-3B=2 \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ $-3 \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$
$= \begin{bmatrix} 4 &8 \\ 6 &4 \end{bmatrix}-\begin{bmatrix} 3 &9 \\ -6 & 15 \end{bmatrix}$
$= \begin{bmatrix} 4-3 &8-9 \\ 6 +6&4-15 \end{bmatrix}$
$= \begin{bmatrix} 1 &-1 \\ 12&-11 \end{bmatrix}$
Note: Here, C is not mandatory

Algebra of Matrices Exercise 4.2 Question 2 (ii)

Answer: $\begin{bmatrix} 9 &-17 \\ -14 &-11 \end{bmatrix}$
Hint: Multiply the C matrix with 4. Then, subtract it from B.
Given:
$A= \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ , $B= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$ and $C= \begin{bmatrix}-2 &5 \\ 3 &4 \end{bmatrix}$
Here, we have to compute$B - 4C.$
Solution:
$B-4C=1 \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$$-4 \begin{bmatrix} -2 &5 \\ 3 &4 \end{bmatrix}$
$= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}-\begin{bmatrix} -8 &20 \\ 12 & 16 \end{bmatrix}$
$= \begin{bmatrix} 1+8 &3-20 \\ -2-12&5-16 \end{bmatrix}$
$= \begin{bmatrix} 9 &-17 \\ -14 &-11 \end{bmatrix}$
Note: Here, A is not mandatory.

Algebra of Matrices Exercise 4.2 Question 2 (iii)

Answer: $\begin{bmatrix} 8 &7 \\ 6 &2 \end{bmatrix}$
Hint: Multiply the A matrix with 3. Then, solve 3A - C.
Given:$A= \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ , $B= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$ and $C= \begin{bmatrix}-2 &5 \\ 3 &4 \end{bmatrix}$
Here, we have to compute $3A - C.$
Solution:
$3A-C=3 \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$$- \begin{bmatrix}- 2 &5 \\ 3 &4 \end{bmatrix}$
$= \begin{bmatrix} 6 &12 \\ 9&6 \end{bmatrix}-\begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$
$= \begin{bmatrix} 6+2 &12-5 \\ 9-3&6-4 \end{bmatrix}$
$= \begin{bmatrix} 8 &7 \\ 6 &2 \end{bmatrix}$
Note: Here, B is not mandatory.

Algebra of Matrices Exercise 4.2 Question 3 (i)

Answer:$\begin{bmatrix} -2 &2 &5 \\ 5& 5 & 1 \end{bmatrix}$
Hint: A + B doesn’t exist because the number of elements in A are not equal to the number of elements in B or we can say, the order of matrices is not equal.
Given: $A= \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ , $B= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$ and $C= \begin{bmatrix}-2 &5 \\ 3 &4 \end{bmatrix}$
Here, we have to compute $A + B$ and $B + C.$
Solution:
$A+B$ doesn’t exist because the no. of rows and columns of both the matrices are not same.
$B + C= \begin{bmatrix} -1 & 0 & 2\\ 3& 4& 1 \end{bmatrix}+\begin{bmatrix} -1 &2 &3 \\ 2 & 1 &0 \end{bmatrix}$
$= \begin{bmatrix} -1-1 & 0+2 & 2+3\\ 3+2 & 4+1 & 1+0 \end{bmatrix}$
$= \begin{bmatrix} -2 & 2 & 5\\ 5 &5 & 1 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 3(ii)

Answer: $\begin{bmatrix} 1 & 6 &1 \\ -6& -13 & -4 \end{bmatrix}$
Hint: $2B + 3A$ don’t exist because the number of elements in A are not equal to the number of elements in B or we can say, the order of matrices is not equal.
Given:$A= \begin{bmatrix} 2 &3 \\ 5 &7 \end{bmatrix}$ , $B= \begin{bmatrix} -1 &0 &2 \\ 3& 4 &1 \end{bmatrix}$ and $C= \begin{bmatrix} -1 & 2& 3\\ 2 &1 &0 \end{bmatrix}$
Here, we have to compute $2B + 3A$ and $3C - 4B.$
Solution:
$2B + 3A$ don’t exist because the no. of rows and columns of both the matrices are not same.
$3C - 4B=3\begin{bmatrix} -1 & 2 & 3\\ 2 & 1 & 0 \end{bmatrix}-4\begin{bmatrix} -1 &0 &2 \\ 3 & 4 &1 \end{bmatrix}$
$=\begin{bmatrix} -3 & 6 & 9\\ 6 & 3 & 0 \end{bmatrix}-\begin{bmatrix} -4 &0 &8 \\ 12 & 16 &4 \end{bmatrix}$
$=\begin{bmatrix} -3+4 & 6-0 & 9-8\\ 6-12 & 3-16 & 0 -4\end{bmatrix}$
$=\begin{bmatrix} 1 & 6 & 1\\ -6 & -13 & -4\end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 3 (iii)

Answer: $\begin{bmatrix} 2 & -14 &-3 \\ 27 & 11 & -11 \end{bmatrix}$
Hint: Just add and subtract the matrix in the form $2A - 3B + 4C.$
Given:$A=\begin{bmatrix} -1 & 0 &2 \\ 3 &1 &4 \end{bmatrix},B=\begin{bmatrix} 0 &-2 &5 \\ 1 & -3 & 1 \end{bmatrix}$ and $C=\begin{bmatrix} 1 & -5 &2 \\ 6 &0 &-4 \end{bmatrix}$
Here, we have to compute $2A - 3B + 4C.$
Solution:
$2A-3B+4C=2\begin{bmatrix} -1 & 0 &2 \\ 3 &1 &4 \end{bmatrix}-3\begin{bmatrix} 0 &-2 &5 \\ 1 & -3 & 1 \end{bmatrix}+4\begin{bmatrix} 1 & -5 &2 \\ 6 &0 &-4 \end{bmatrix}$
$= \begin{bmatrix} -2 &0 &4 \\ 6 & 2 & 8 \end{bmatrix}-\begin{bmatrix} 0 & -6 &15 \\ 3 &-9 & 3 \end{bmatrix}+\begin{bmatrix} 4 & -20 &8 \\ 24 & 0 & -16 \end{bmatrix}$
$= \begin{bmatrix} -2-0+4 &0+6-20 &4-15+8 \\ 6-3+24 &2+9+0 & 8-3-16 \end{bmatrix}$
$= \begin{bmatrix} 2 & -14 &-3 \\ 27 & 11 & -11 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 2 (iv)

Answer:$\begin{bmatrix} -2 &21 \\ 22 &8 \end{bmatrix}$
Hint: Multiply the A, B, and C with 3, 2, and 3 respectively. Then, solve 3A – 2B + 3C.
Given: $A= \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$ , $B= \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}$ and $C= \begin{bmatrix}-2 &5 \\ 3 &4 \end{bmatrix}$
Here, we have to compute $3A - 2B + 3C.$
Solution:
$3A-2B+3C=3 \begin{bmatrix} 2 &4 \\ 3 &2 \end{bmatrix}$$-2 \begin{bmatrix} 1 &3 \\ -2 &5 \end{bmatrix}+3\begin{bmatrix} -2 & 5\\ 3 & 4 \end{bmatrix}$
$= \begin{bmatrix} 6 & 12\\ 9 &6 \end{bmatrix}- \begin{bmatrix} 2 &6 \\ -4 &10 \end{bmatrix}+\begin{bmatrix} -6 & 15\\ 9& 12 \end{bmatrix}$
$= \begin{bmatrix} 6-2-6 & 12-6+15\\ 9+4+9 &6-10+12 \end{bmatrix}$
$= \begin{bmatrix} -2 &21 \\ 22 &8 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 5 (i)

Answer:$diag (0, -7, 17)$
Hint:
$diag=\begin{bmatrix} 1 &0 &0 \\ 0& 1&0 \\ 0& 0 & 1 \end{bmatrix}$ using this formula in $A, B, C$ and solve $A-2B.$
Given: $A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)$
Here, we have to compute:$A-2B.$
Solution:
$A=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix},B=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix}$
$A-2B=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix}-2\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix}$
$=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix}-2\begin{bmatrix} 2 & 0 &0 \\ 0 & 2 &0 \\ 0 &0 & -8 \end{bmatrix}$
$=\begin{bmatrix} 2 -2&0 -0 &0 -0\\ 0- 0& -5-2 &0-0 \\ 0-0 & 0 -0& 9+8 \end{bmatrix}$
$=\begin{bmatrix} 0&0&0\\ 0& -7 &0 \\ 0 & 0 & 17 \end{bmatrix}$
$= diag (0, -7, 17)$
Note: Here, C is not mandatory.

Algebra of Matrices Exercise 4.2 Question 5 (ii)

Answer:$diag (-9 14 -18)$
Hint:
$diag= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1 \end{bmatrix}$ using this formula in $A, B, C$ and solve $B + C -2A.$
Given:$A = diag (2 -5 9), B = diag (1 1 -4) and\: C = diag (-6 3 4)$
Here, we have to compute: $B + C -2A.$
Solution:
$A=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix},B=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix},C= \begin{bmatrix} -6 &0 & 0\\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}$
$B+C-2A=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0 & 0 & -4 \end{bmatrix}+\begin{bmatrix} -6 & 0 &0 \\ 0 & 3 &0 \\ 0 &0 & 4 \end{bmatrix}-2 \begin{bmatrix} 2 &0 & 0\\ 0 & -5 &0 \\ 0 & 0 & 9 \end{bmatrix}$
$=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0 & 0 & -4 \end{bmatrix}+\begin{bmatrix} -6 & 0 &0 \\ 0 & 3 &0 \\ 0 &0 & 4 \end{bmatrix}-\begin{bmatrix} 4 &0 & 0\\ 0 & -10 &0 \\ 0 & 0 & 18 \end{bmatrix}$
$=\begin{bmatrix} 1-6-4 &0+0-0 &0+0-0 \\ 0+0-0& 1+3+10 &0+0-0 \\ 0 +0-0& 0+0-0 & -4+4-18 \end{bmatrix}$
$=\begin{bmatrix} -9 &0 &0\\ 0& 14 &0 \\ 0 & 0 & -18 \end{bmatrix}$
$= diag (-9\: \ 14 -18)$

Algebra of Matrices Exercise 4.2 Question 5 (iii)

Answer:$diag (37 -22 -14)$
Hint:

using this formula in$A, B, C$ and solve $2A + 3B - 5C.$
Given:$A = diag (2 -5 9), B = diag (1 1 -4) and\: C = diag (-6 3 4)$
Here, we have to compute:$2A + 3B - 5C.$
Solution:
$A=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix},B=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix},C= \begin{bmatrix} -6 &0 & 0\\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}$
$2A+3B-5C=2\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix}+3\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix}-5 \begin{bmatrix} -6 &0 & 0\\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}$
$=\begin{bmatrix} 4 &0 &0 \\ 0& -10 &0 \\ 0 & 0 & 18 \end{bmatrix}+3\begin{bmatrix} 3 & 0 &0 \\ 0 & 3 &0 \\ 0 &0 & -12 \end{bmatrix}-5 \begin{bmatrix} -30 &0 & 0\\ 0 & 15 &0 \\ 0 & 0 & 20 \end{bmatrix}$
$=\begin{bmatrix} 4+3+30 &0+0-0 &0+0-0 \\ 0+0-0& -10+3-15 &0+0-0 \\ 0+0-0 & 0+0-0 & 18-12-20 \end{bmatrix}$
$=\begin{bmatrix} 37 &0 &0\\ 0& -22 &0\\ 0 & 0 & -14 \end{bmatrix}$
$= diag (37 -22 -14)$

Algebra of Matrices Exercise 4.2 Question 6

Hint: In LHS, solve (A + B) by adding A and B matrix. Then, add C matrix. and In RHS, first solve (B+C) then add A to it.
Given: To prove:$(A + B) + C = A + (B + C)$
$A = \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} , B = \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} , C = \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$
Solution:
$LHS = (A + B) + C$
$= \left ( \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} + \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} \right )+ \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$
$= \left ( \begin{bmatrix} 2+9 & 1+7 & 1-1\\ 3+3 & -1+5& 0+4\\ 0+2 & 2+1 & 4+6 \end{bmatrix} \right )+ \begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix}$
$= \begin{bmatrix} 2+9+2 & 1+7 -4& 1-1+3\\ 3+3+1 & -1+5-1& 0+4+0\\ 0+2+9 & 2+1 +4& 4+6 +5\end{bmatrix}$
$= \begin{bmatrix} 13 & 4& 3\\ 7 & 3& 4\\ 11 & 7& 15\end{bmatrix}$ … (1)

$RHS = A + (B + C)$
$= \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} \left ( \begin{bmatrix} 9& 7&-1 \\ 3 & 5 & 4\\ 2 & 1 &6 \end{bmatrix} +\begin{bmatrix} 2 & -4 & 3\\ 1 & -1 &0 \\ 9 &4 & 5 \end{bmatrix} \right )$
$= \begin{bmatrix} 2 & 1 & 1\\ 3 & -1& 0\\ 0 & 2 & 4 \end{bmatrix} \left ( \begin{bmatrix} 9+2& 7-4&-1+3 \\ 3+1 & 5-1 & 4+0\\ 2+9 & 1+4 &6 +5\end{bmatrix} \right )$
$= \begin{bmatrix} 2+9+2&1+ 7-4&1-1+3 \\3+ 3+1 &-1+0+ 5-1 & 0+4+0\\0+ 2+9 & 2+1+4 &4+6 +5\end{bmatrix}$
$= \begin{bmatrix} 13&4&3 \\7 &3 & 4\\11 & 7 &15\end{bmatrix}$ … (2)
From (1) and (2)
LHS = RHS
Hence,$(A + B) + C = A + (B + C)$ is proved.

Algebra of Matrices Exercise 4.2 Question 7

Answer:$X= \begin{bmatrix} 4 & 4\\ 0 & 4 \end{bmatrix} and \: Y=\begin{bmatrix} 1 &-2 \\ 0 & 5 \end{bmatrix}$
Hint: Try to add two matrices and find variable ‘X ’. Then substitute ‘X ’ value in any matrix to find ‘Y ’.
Given: $X + Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix} and\: X - Y = \begin{bmatrix} 3&6 \\ 0 & -1 \end{bmatrix}$
Here, we have to compute X and Y.
Solution:
$\left ( X + Y \right )+\left ( X - Y \right ) = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3&6 \\ 0 & -1 \end{bmatrix}$
$\left ( X + Y \right )+\left ( X - Y \right ) = \begin{bmatrix} 5+3 & 2+6\\ 0+0 & 9-1 \end{bmatrix}$
$2X = \begin{bmatrix} 8 & 8\\ 0 & 8 \end{bmatrix}$
$X =\frac{1}{2} \begin{bmatrix} 8 & 8\\ 0 & 8 \end{bmatrix}$
$X = \begin{bmatrix} 4 & 4\\ 0 & 4 \end{bmatrix}$
Now,
$X+Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}$
$\begin{bmatrix} 4 &4 \\ 0 & 4 \end{bmatrix}+Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}$
$Y = \begin{bmatrix} 5 & 2\\ 0 & 9 \end{bmatrix}-\begin{bmatrix} 4 &4 \\ 0 & 4 \end{bmatrix}$
$Y = \begin{bmatrix} 1&- 2\\ 0 & 5 \end{bmatrix}$
$X= \begin{bmatrix} 4 & 4\\ 0 & 4 \end{bmatrix} and \: Y=\begin{bmatrix} 1 &-2 \\ 0 & 5 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 8

Answer: $X= \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$
Hint: Substitute $Y value\: in\: 2X+Y .$
Given:$2X+Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix} and \: Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$
Here, we have to compute X .
Solution:
$2X+Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}, Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$
$2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$
$2X= \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$
$X= \frac{1}{2}\begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}$
$X= \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$

Answer: X

= 3-21-21-1

and Y

= 02200-3

Hint: Try to take the one variable out by using Addition and Multiplication.
Given: 2X-Y

= 6-60-421

and X+2Y

= 325-21-7

Here, we have to compute X

and Y

.
Solution:
2X-Y

=6-60-421

… (1)
X+2Y

= 325-21-7

… (2)
Multiplying equation (1) with 2, we get:
2(2X-Y

) = 26-60-421

4X-2

Y = 12-120-842

… (3)
Adding equation (2) with equation (3), we get:
(4X-2

Y) + (X+2Y

) = 12-120-842

+ 325-21-7

4X-2

Y + X+2Y

= 12+3-12+20+5-8-24+12-7

= 15-105-105-5

= 15

15-105-105-5

= 3-21-21-1

Applying ‘X

’ value in equation (2), we get:
X+2Y

= 325-21-7

3-21-21-1

+ 2Y

= 325-21-7

= 3-32+25-1-2+21-1- 7+1

= 12

04400-6

= 02200-3

Algebra of Matrices Exercise 4.2 Question 10

Answer: $X = \begin{bmatrix} 2 &3 &1 \\ 0 &1 &2 \\ 6 &4 &0 \end{bmatrix} and Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix}$
Hint: Try to add two matrices and find variable ‘X ’. Then substitute ‘X ’ value in any matrix to find ‘Y ’.
Given:$X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix} and X-Y = \begin{bmatrix} 1 &1 &1 \\ 1 &1 &0 \\ 1&0 &0 \end{bmatrix}$
Here, we have to compute X and Y .
Solution:
$X+Y+X-Y = \begin{bmatrix} 3 & 5 &1 \\ -1 &1 &4 \\ 11 & 8 &0 \end{bmatrix} +\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 0\\ 1 &0 & 0 \end{bmatrix}$
$X+Y+X-Y= \begin{bmatrix} 3+1 & 5+1& 1+1\\ -1+1 &1+1 &4+0 \\ 11+1 & 8+0 & 0+0 \end{bmatrix}$
$2X= \begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix}$
$X= \frac{1}{2}\begin{bmatrix} 4 & 6& 2\\ 0 &2 &4\\ 12 & 8 & 0 \end{bmatrix}$
$X= \begin{bmatrix} 2 & 3& 1\\ 0 &1 &2\\ 6 & 4 & 0 \end{bmatrix}$
Now,
$X+Y= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$
$\begin{bmatrix} 2 & 3& 1\\ 0 &1 &2\\ 6 & 4 & 0 \end{bmatrix}+Y$ $= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$$= \begin{bmatrix} 3 &5 &1 \\ -1 &1 &4 \\ 11 & 8 & 0 \end{bmatrix}$
$Y = \begin{bmatrix} 3-2 & 5-3 & 1-1\\ -1-0 & 1-1 &4-2 \\ 11-6 &8-4 & 0-0 \end{bmatrix}$
$Y = \begin{bmatrix} 1 & 2 &0 \\ -1 & 0 &2 \\ 5 & 4 & 0 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 11

Answer:$A = \begin{bmatrix} 8 & -3& 5\\ -2 & -3 & -6 \end{bmatrix}$
Hint: Transposing the matrix from LHS to RHS
Given:$\begin{bmatrix} 1 & 2 &-1 \\ 0 & 4 & 9 \end{bmatrix} + A =\begin{bmatrix} 9 &-1 &4 \\ -2& 1 &3 \end{bmatrix}$
Here, we have to compute A.
Solution:
$\begin{bmatrix} 1 & 2 &-1 \\ 0 & 4 & 9 \end{bmatrix} + A =\begin{bmatrix} 9 &-1 &4 \\ -2& 1 &3 \end{bmatrix}$
$A =\begin{bmatrix} 9 &-1 &4 \\ -2& 1 &3 \end{bmatrix}-\begin{bmatrix} 1 & 2 &-1 \\ 0 & 4 & 9 \end{bmatrix}$
$A =\begin{bmatrix} 9-1 &-1-2 &4 +1\\ -2-0& 1-4 &3 -9 \end{bmatrix}$
$A = \begin{bmatrix} 8 & -3& 5\\ -2 & -3 & -6 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 12

Answer: $C = \begin{bmatrix} -24 &-10 \\ -28 & -38 \end{bmatrix}$
Hint: Null matrix is $\begin{bmatrix} 0 &0\\ 0 & 0 \end{bmatrix}$
Given: $A =\begin{bmatrix} 9 &1 \\ 7 & 8 \end{bmatrix} , B =\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix}, 5A + 3B + 2C = 0$
Here, we have to compute C.
Solution:
$5A + 3B + 2C = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$
$5\begin{bmatrix} 9 & 1\\ 7 & 8 \end{bmatrix} + 3\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$
$\begin{bmatrix} 45 & 5\\ 35 & 40 \end{bmatrix} + \begin{bmatrix} 3 & 15\\ 21 & 36 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$
$\begin{bmatrix} 45+3 & 5+15\\ 35+21 & 40+36 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$
$\begin{bmatrix} 48 & 20\\ 56 & 76 \end{bmatrix} + 2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}$
$2C = \begin{bmatrix} 0 &0 \\ 0&0 \end{bmatrix}-\begin{bmatrix} 48 & 20\\ 56 & 76 \end{bmatrix}$
$2C = \begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}$
$C =\frac{1}{2}\begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}$
$C = \begin{bmatrix} -24 &-10 \\ -28 & -38 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 13

Answer:
$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$
Hint: Try to separate the ‘X’ variable into LHS.
Given:$A =\begin{bmatrix} 2 &-2 \\ 4 &2 \\ -5 & 1 \end{bmatrix} , B =\begin{bmatrix} 8 & 0\\ 4 & -2\\ 3 & 6 \end{bmatrix} , 2A + 3x = 5B$
Here, we have to compute x.
Solution:
$2A + 3 X = 5B$
$2\begin{bmatrix} 2 & -2\\ 4& 2\\ -5& 1 \end{bmatrix}+3X=5\begin{bmatrix} 8 &0 \\ 4& -2\\ 3 & 6 \end{bmatrix}$
$\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}+3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}$
$3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}-\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}$
$3X=\begin{bmatrix} 40-4 &0+4 \\ 20-8& -10-4\\ 15+10 & 30-2 \end{bmatrix}$
$3X=\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$
$X=\frac{1}{3}\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$
$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 14

Answer: $C = \begin{bmatrix} -3 &4 &-1 \\ -3& 0 &-1 \end{bmatrix}$
Hint: Zero matrix is $\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$
Add A and B matrices. Then, find C.
Given:$A =\begin{bmatrix} 1 &-3 &2 \\ 2 & 0 &2 \end{bmatrix} , B =\begin{bmatrix} 2 &-1 & -1\\ 1& 0 & -1 \end{bmatrix} , A + B + C = 0$
Here, we have to compute C.
Solution:
$A + B + C =\begin{bmatrix} 0 &0 &0 \\ 0& 0 & 0 \end{bmatrix}$
$\begin{bmatrix} 1 & -3 &2 \\ 2 &0 &2 \end{bmatrix} +\begin{bmatrix} 2 & -1&-1 \\ 1& 0 & -1 \end{bmatrix} + C = \begin{bmatrix} 0 &0 &0 \\ 0& 0& 0 \end{bmatrix}$
$\begin{bmatrix} 3 & -4 &1\\ 3 &0 &1 \end{bmatrix} + C = \begin{bmatrix} 0 &0 &0 \\ 0& 0& 0 \end{bmatrix}$
$C = \begin{bmatrix} 0 &0 &0 \\ 0& 0& 0 \end{bmatrix}-\begin{bmatrix} 3 & -4 &1\\ 3 &0 &1 \end{bmatrix}$
$C = \begin{bmatrix} -3 &4 &-1 \\ -3& 0 &-1 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question15 (i)

Answer:$x = 32\: and\: y = -32$
Hint: Solve LHS and then equate with RHS.
Given:$\begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
Here, we have to compute x and y.
Solution:
$\begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
$\begin{bmatrix} x-y+3 &2-2 & -2+2\\ 4+1 &x+0 & 6-1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
Equating this, we get:
$x-y + 3 = 6$
$\Rightarrow x-y = 3$ … (1)
Also,
$x = 2x + y$
$\Rightarrow y = -x$ …. (2)
Applying the value of ‘y’ in equation (1), we get:
$x-(-x) = 3$
$2x = 3$
$x = 32$
Applying the value of ‘x ’ in equation (2), we get:$y = -32$

Algebra of Matrices Exercise 4.2 Question 15 (ii)

Answer: $x = 1, y = 3\: and\: z = 10$
Hint: Solve LHS and then equate with RHS.
Given: $\begin{bmatrix} x & y+z & z-3 \end{bmatrix}+\begin{bmatrix} y & 4 & 5 \end{bmatrix}= \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$
Here, we have to compute x , y and z.
Solution:
$\begin{bmatrix} x & y+2 & z-3 \end{bmatrix}+\begin{bmatrix} y & 4 & 5 \end{bmatrix}= \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$
$\begin{bmatrix} x+y & y+2+4 & z-3+5 \end{bmatrix}= \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$
Equating this, we get:
$x+y = 4$ …(1)
Also,
$y + 6 = 9$
$\Rightarrow y = 3$
$z + 2 = 12$
$\Rightarrow z = 10$
Applying the value of ‘y’ in equation (1), we get:
$x + 3 = 4$
$x = 1$

Algebra of Matrices Exercise 4.2 Question 15 (iii)

Answer:$x = 1, y = 2$
Hint: Solve LHS part and then equate with RHS part.
Given:
$x \begin{bmatrix} 2\\1 \end{bmatrix} + y \begin{bmatrix} 3\\5 \end{bmatrix} +\begin{bmatrix} -8\\11 \end{bmatrix} = 0$
Here, we have to compute x and y .
Solution:
$x \begin{bmatrix} 2\\1 \end{bmatrix} + y \begin{bmatrix} 3\\5 \end{bmatrix} +\begin{bmatrix} -8\\11 \end{bmatrix} = 0$
$\begin{bmatrix} 2x+3y-8\\x+5y-11 \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix}$
Equating this, we get:
$2x+3y-8 = 0$
$\Rightarrow 2x+3y=8$ …(1)
Also,
$x+5y-11 = 0$
$\Rightarrow x=11-5y$ …(2)
Applying the value of ‘x

’ in equation (1), we get:
$2(11 -5y) + 3y = 8$
$22 -10y + 3y = 8$
$-7y = 8 -22$
$-7y = -14$
$y = 2$
Applying the value of ‘y’ in equation (2), we get:
$x = 11 - 5(2)$
$x = 11- 10$
$x = 1$

Algebra of Matrices Exercise 4.2 Question 16

Answer:$x = 2, y = -8$
Hint: Add LHS part and then equate with RHS part.
Given:$2\begin{bmatrix} 3 & 4\\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 &y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 &0 \\ 10& 5 \end{bmatrix}$
Here, we have to compute x and y.
Solution:
$2\begin{bmatrix} 3 & 4\\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 &y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 &0 \\ 10& 5 \end{bmatrix}$
$\begin{bmatrix} 6 & 8\\ 10 &2 x \end{bmatrix} + \begin{bmatrix} 1 &y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 &0 \\ 10& 5 \end{bmatrix}$
$\begin{bmatrix} 6+1 & 8+y\\ 10+0 &2 x+1 \end{bmatrix} = \begin{bmatrix} 7 &0 \\ 10& 5 \end{bmatrix}$
$\begin{bmatrix} 7 & 8+y\\ 10 &2 x+1 \end{bmatrix} = \begin{bmatrix} 7 &0 \\ 10& 5 \end{bmatrix}$
Equating this, we get:
$8+y = 0$
$\Rightarrow y= -8$
Also,
$2x +1 = 5$
$2x = 4$
$\Rightarrow x =2$

Algebra of Matrices Exercise 4.2 Question 17

Answer:$\lambda = 2$
Hint: Add LHS part of matrix and separate the variable $\lambda$ .Then, solve.
Given:$\lambda \begin{bmatrix} 1 & 0 &2 \\ 3 & 4 & 5 \end{bmatrix} +2 \begin{bmatrix} 1 & 2 & 3\\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 &10 \\ 4 & 2 & 14 \end{bmatrix}$
Here, we have to find the value of $\lambda$
Solution:
$\lambda \begin{bmatrix} 1 & 0 &2 \\ 3 & 4 & 5 \end{bmatrix} +2 \begin{bmatrix} 1 & 2 & 3\\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 &10 \\ 4 & 2 & 14 \end{bmatrix}$
$\lambda \begin{bmatrix} \lambda & 0 &2\lambda \\ 3\lambda & 4\lambda & 5\lambda \end{bmatrix} +2 \begin{bmatrix} 1 & 2 & 3\\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 &10 \\ 4 & 2 & 14 \end{bmatrix}$
$\lambda \begin{bmatrix} \lambda+2 & 0+4 &2\lambda +6 \\ 3\lambda -2 & 4\lambda -6 & 5\lambda +4 \end{bmatrix} = \begin{bmatrix} 4 & 4 &10 \\ 4 & 2 & 14 \end{bmatrix}$
Equating this, we get:
$\lambda +2 = 4$
$\lambda = 4 - 2$
$\Rightarrow \lambda = 2$

Algebra of Matrices Exercise 4.2 Question 18 (i)

Answer:$X = \begin{bmatrix} -1 & -2\\ -7 & -13 \end{bmatrix}$
Hint: Add $2A + B$ matrix then transpose to RHS.
Given:$A = \begin{bmatrix} -1 &2 \\ 3 & 4 \end{bmatrix} , B =\begin{bmatrix} 3 & -2\\ 1 & 5 \end{bmatrix}, 2A + B + x =0$
Here, we have to compute X .
Solution:
$2A + B + X =0$
$2\begin{bmatrix} -1 & 2\\ 3 & 4 \end{bmatrix} +\begin{bmatrix} 3 &-2 \\ 1 & 5 \end{bmatrix} + X = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$
$\begin{bmatrix} -2 & 4\\ 6 & 8 \end{bmatrix} +\begin{bmatrix} 3 &-2 \\ 1 & 5 \end{bmatrix} + X = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$
$\begin{bmatrix} -2+3 & 4-2\\ 6+1 & 8+5 \end{bmatrix} + X = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$
$\begin{bmatrix} 1 & 2\\ 7 & 13 \end{bmatrix} + X = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$$X = \begin{bmatrix} -1 & -2\\ -7 & -13 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 18 (ii)

Answer:$X = \begin{bmatrix} -2 &\frac{-10}{3} \\ \\ 12 &\frac{14}{3} \\ \\ \frac{-31}{3} & \frac{-7}{3} \end{bmatrix}$
Hint: Take the variable 3X into RHS and 5B into LHS. Then, solve.
Given:$A =\begin{bmatrix} 8 & 0\\ 4&-2 \\ 3&6 \end{bmatrix} , B =\begin{bmatrix} 2 &-2 \\ 4& 2\\ -5& 1 \end{bmatrix} , 2A + 3X = 5B$
Here, we have to compute X.
Solution:
$2A + 3X = 5B$
$2\begin{bmatrix} 8 & 0\\ 4& -2\\ 3&6 \end{bmatrix} + 3X = 5\begin{bmatrix} 2&-2 \\ 4 &2 \\ -5 &1 \end{bmatrix}$
$\begin{bmatrix} 16 & 0\\ 8& -4\\ 6&12 \end{bmatrix} + 3X = \begin{bmatrix} 10&-10 \\ 20 &10 \\ -25 &5 \end{bmatrix}$
$3X = \begin{bmatrix} 10&-10 \\ 20 &10 \\ -25 &5 \end{bmatrix}-\begin{bmatrix} 16 & 0\\ 8& -4\\ 6&12 \end{bmatrix}$
$3X = \begin{bmatrix} -6 &-10 \\ 12 &14 \\ -31 &-7 \end{bmatrix}$
$X = \frac{1}{3}\begin{bmatrix} -6 &-10 \\ 12 &14 \\ -31 &-7 \end{bmatrix}$
$X = \begin{bmatrix} -2 &\frac{-10}{3} \\ \\ 12 &\frac{14}{3} \\ \\ \frac{-31}{3} & \frac{-7}{3} \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 19 (i)

Answer:$x = 2, y = 4, z = 1, t =3$
Hint: Solve LHS part and equate with RHS part.
Given: $3 \begin{bmatrix} x &y \\ z&t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$
Here, we have to compute x, y, z and t.
Solution:
$3 \begin{bmatrix} x &y \\ z&t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$
$\begin{bmatrix} 3x &3y \\ 3z&3t \end{bmatrix} = \begin{bmatrix} x &6 \\ -1 &2t \end{bmatrix} + \begin{bmatrix} 4 &x+y \\ z+t & 2t+3 \end{bmatrix}$
$\begin{bmatrix} 3x &3y \\ 3z&3t \end{bmatrix} = \begin{bmatrix} x+4 &6+x+y \\ -1+z+t &2t+3 \end{bmatrix}$
Equating this, we get:
$\! \! \! \! \! \! \! \! 3 x = x + 4\\ 2x = 4 \\ \Rightarrow x = 2$
Also,
$\! \! \! \! \! \! \! \! 3y = 6 + x + y \\ 2y = 6 + x$
Putting the value of ‘x ’, we get:
$\! \! \! \! \! \! \! \! 2y = 6 + 2\\ 2y = 8\\ \Rightarrow y = 4$
Also,
$\! \! \! \! \! \! \! 3t = 2t + 3\\ \Rightarrow t = 3$
Now,
$\! \! \! \! \! \! \! \! \! 3z = -1 + z + t\\ 2z = - 1 + t$
Putting the value of ‘t’, we get:
$\! \! \! \! \! \! \! 2z = -1 + 3\\ 2z = 2\\ \Rightarrow z = 1$

Algebra of Matrices Exercise 4.2 Question 19 (ii)

Answer:$x = 2\: and\: y = 9$
Hint: Add LHS part and equate with RHS part.
Given:$2 \begin{bmatrix} x & 5\\ 7& y-3 \end{bmatrix} + \begin{bmatrix} 3 &4 \\ 1 &2 \end{bmatrix} = \begin{bmatrix} 7 & 14\\ 15 & 14 \end{bmatrix}$
Here, we have to compute x, y, z and t.
Solution:
$2 \begin{bmatrix} x & 5\\ 7& y-3 \end{bmatrix} + \begin{bmatrix} 3 &4 \\ 1 &2 \end{bmatrix} = \begin{bmatrix} 7 & 14\\ 15 & 14 \end{bmatrix}$
$\begin{bmatrix} 2x & 10\\ 14&2 y-6 \end{bmatrix} + \begin{bmatrix} 3 &4 \\ 1 &2 \end{bmatrix} = \begin{bmatrix} 7 & 14\\ 15 & 14 \end{bmatrix}$
$\begin{bmatrix} 2x+3 & 10+4\\ 14+1&2 y-6+2 \end{bmatrix} = \begin{bmatrix} 7 & 14\\ 15 & 14 \end{bmatrix}$
Equating this, we get:
$\! \! \! \! \! \! \! \! \! 2 x + 3 = 7\\ 2x = 4\\ \Rightarrow x = 2$
Also,
$\! \! \! \! \! \! \! \! \! \! 2y - 6 + 2 = 14\\ 2y -4 = 14\\ 2y = 18\\\Rightarrow y=9$

Algebra of Matrices Exercise 4.2 Question 20

Answer:$X = \begin{bmatrix} -2 &0 \\ -1 & -3 \end{bmatrix} and\: Y = \begin{bmatrix} 2& 1\\ 2 & 2 \end{bmatrix}$
Hint: Subtract matrix,$2X + 3Y and\: 3X + 2Y$. Then, solve.
Given:$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} and\: 3X + 2Y = \begin{bmatrix} -2 &2 \\ 1 & -5 \end{bmatrix}$
Here, we have to compute x and y.
Solution:
$3 (2X + 3Y) - 2 (3X + 2Y) = 3 \begin{bmatrix} 2 & 3\\ 4& 0 \end{bmatrix} - 2 \begin{bmatrix} -2 &2 \\ 1&-5 \end{bmatrix}$
$6X + 9Y - 6X - 4Y =\begin{bmatrix} 6 &9 \\ 12 & 0 \end{bmatrix} - \begin{bmatrix} -4 & 4\\ 2& -10 \end{bmatrix}$
$5Y = \begin{bmatrix} 6+4 & 9-4\\ 12-2 & 0+10 \end{bmatrix}$
$5Y = \begin{bmatrix} 10 &5 \\ 10 & 10 \end{bmatrix}$
$Y = \begin{bmatrix} 2 &1\\ 2 & 2 \end{bmatrix}$
Also,
$2 (2X + 3Y) - 3 (3X + 2Y) = 2 \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - 3 \begin{bmatrix} -2 &2 \\ 1 & -5 \end{bmatrix}$
$4X + 6Y -9X-6Y =\begin{bmatrix} 4 &6 \\ 8 & 0 \end{bmatrix} - \begin{bmatrix} -6 &6 \\ 3 & -15 \end{bmatrix}$
$-5X = \begin{bmatrix} 4+6 &6-6 \\ 8-3 & 0+15 \end{bmatrix}$
$-5X = \begin{bmatrix} 10 &0 \\ 5 & 15 \end{bmatrix}$
$X = \begin{bmatrix} -2 &0 \\ -1& -3 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 21

Answer:
$\begin{bmatrix} 450\\ 180\\ 30\\ 30 \end{bmatrix}$
Hint: Put a matrix for peons, clerks, typist, sector officers. Multiply with 30 for calculating all 30 colleges.
Given: There are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 sector officer.
Solution:
No. of different types of posts in any college is given by:
$X=\begin{bmatrix} 15\\ 6\\ 1\\ 1 \end{bmatrix}$
Total no. of posts of each kind in all the colleges = 30X
$=30\begin{bmatrix} 15\\ 6\\ 1\\ 1 \end{bmatrix}$
$=\begin{bmatrix} 450\\ 180\\ 30\\ 30 \end{bmatrix}$

Algebra of Matrices Exercise 4.2 Question 22

Answer:$X = 30000\: and \: Y = 15000$
Hint: Try to solve monthly saving of Aryan and Babban using the ratio.
Recall that the solution to the system of equations that can be written as $AX = B.$$AX = B.$
Given: The monthly income of Aryan and Babban are in the ratio $3:4.$The monthly expenditures are in the ratio$5:7.$Saving of each month is $Rs. 15000.$
Solution:
Let the monthly income of Aryan and Babban be $3X$ and $4X$ respectively. Suppose that monthly expenditures are $5Y$ and$7Y$ respectively.
Since, each save $Rs. 15000$ per month,
Monthly saving of Aryan:$3X - 5Y = 15000$
Monthly saving of Babban:$4X - 7Y = 15000$
The above system of equations can be written in the matrix form as follows:
$\begin{bmatrix} 3 &-5 \\ 4 & -7 \end{bmatrix}\begin{bmatrix} X\\Y \end{bmatrix}= \begin{bmatrix} 15000\\15000 \end{bmatrix}$
Now,
$A = \begin{bmatrix} 3 & -5\\ 4 & -7 \end{bmatrix} = 3\times \left ( -7 \right )-4\times \left ( -5 \right )$
$= -21 - (-20)$
$= -1$
Adj$(A) = \begin{bmatrix} -7 &-4 \\ 5 & 3 \end{bmatrix}^{T} = \begin{bmatrix} -7 &5 \\ -4& 3 \end{bmatrix}$
So,
$A^{-1} = \frac{1}{\left | A \right |} Adj (A)$
$= (-1) \begin{bmatrix} - 7& 5\\ -4& 3 \end{bmatrix}$
$= \begin{bmatrix} 7& -5\\ 4& -3 \end{bmatrix}$
$\therefore X = A^{-1}B$
$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 7 & -5\\ 4 & -3 \end{bmatrix}\begin{bmatrix} 15000\\15000 \end{bmatrix}$
$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 105000 &-75000 \\ 60000 & -45000 \end{bmatrix}$
$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 30000\\15000 \end{bmatrix}$
$X = 30000, Y = 15000$
Therefore,
Monthly income of Aryan $= 3X$
$\! \! \! \! \! \! \! \! \! = 3 (30000)\\ = Rs. 90,000$
Monthly income of Aryan$= 4X$
$\! \! \! \! \! \! \! \! = 4 (30000)\\ = Rs. 1,20,000$
Note:
From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.

Algebra of Matrices Exercise 4.2 Question 15 (i)

Answer:$x = \frac{3}{2}\: and\: y = \frac{-3}{2}$
Hint: Solve LHS and then equate with RHS.
Given:$\begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 &- 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
Here, we have to compute x and y.
Solution:
$\begin{bmatrix} x-y &2 & -2\\ 4 &x & 6 \end{bmatrix} +\begin{bmatrix} 3 & -2 &2 \\ 1 & 0 &- 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
$\begin{bmatrix} x-y+3 &2-2 & -2+2\\ 4+1 &x+0 & 6-1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5 \end{bmatrix}$
Equating this, we get:
$x-y + 3 = 6$
$\Rightarrow x-y = 3$ … (1)
Also,
$x = 2x + y$
$\Rightarrow y = -x$ …. (2)
Applying the value of ‘y’ in equation (1), we get:
$x-(-x) = 3$
$2x = 3$
$x = \frac{3}{2}$
Applying the value of ‘x ’ in equation (2), we get:$y = \frac{-3}{2}$

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