NCERT Solutions for Class 12 Maths Chapter 3 Matrices are provided here. Matrices is an important and powerful tool in mathematics and it's basically introduced to solve simultaneous linear equations. maths chapter 3 class 12 includes the type of matrices, operations of matrices, and many more concepts. These concepts are used to solve challenging problems of 3d geometry, and determinants Therefore matrices class 12 solutions become important to get command of concepts. The NCERT Solutions are prepared by expert team according the latest syllabus of CBSE 2023. Also, Students can refer matrices class 12 NCERT solutions which covers all the questions of NCERT Books for Class 12 Maths.

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Matrices Class 12 Questions And Answers
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NCERT Solutions for Class 12 Maths Chapter 3 Matrices - Important Formulae
Class 12 Maths Chapter 3 Question Answer (Intext Questions and Exercise)
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NCERT Solutions for Class 12 Maths Chapter 3 Matrices

The practice from NCERT solutions for class 12 maths Chapter 3 Matrices is very important to score well in academics as well as in competitive exams. Matrices class 12 also includes exercises you can also solve to build hold on the concepts. All the questions in NCERT maths class 12 chapter 3 are explained in a detailed manner. You can also refer to class 12 maths ncert solutions chapter 3 matrices by clicking on the link NCERT solutions for Class 12.

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Matrices Class 12 Questions And Answers PDF Free Download

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NCERT Solutions for Class 12 Maths Chapter 3 Matrices - Important Formulae

Matrix Definition and Properties:

A matrix is an ordered rectangular array of numbers or functions.

A matrix of order m × n consists of m rows and n columns.

The order of a matrix is written as m × n, where m is the number of rows and n is the number of columns.

A matrix is called a square matrix when m = n.

A diagonal matrix A = [a_ij]_m×m has a_ij = 0 when i ≠ j.

A scalar matrix A = [a_ij]_n×n has a_ij = 0 when i ≠ j, aij = k (where k is a constant)

when i = j.

An identity matrix A = [a_ij]n×n has a_ij = 1 when i = j and a_ij = 0 when i ≠ j.

A zero matrix contains all its elements as zero.

A column matrix is of the form [A]_{n × 1}.

A row matrix is of the form [A]_{1 × n}.

Equality of Matrices:

Two matrices A and B are equal (A = B) if they have the same order and a_ij = b_ij for all the corresponding values of i and j.

Operations on Matrices:

Matrix Addition:

If A = [a_ij]_{m × n} and B = [b_ij]_{m × n}, then A + B = [a_ij + b_ij]m × n.

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Matrix Subtraction:

If A = [a_ij]_{m × n} and B = [b_ij]_{m × n}, then A - B = [a_ij - b_ij]_{m × n}.

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Multiplication of a Matrix by Scalar:

Let A = [a_ij]_{m × n} be a matrix and k is a scalar, then kA is obtained by multiplying each element of A by the scalar k, i.e., kA = [ka_ij]_{m × n}.

Multiplication of Matrices:

Let A be an m × p matrix, and B be a p × n matrix. Their product AB is defined if the number of columns in A is equal to the number of rows in B. The resulting matrix is an m × n matrix, and the elements are calculated as follows: (AB)_ij = Σ(a_i * b_j), where the sum is taken over all values of p.

Transpose of a Matrix:

The transpose of a matrix A, denoted as AT, is obtained by interchanging its rows and columns.

Symmetric and Skew-Symmetric Matrices:

A matrix A is symmetric if A = AT (i.e., it is equal to its transpose).

A matrix A is skew-symmetric if AT = -A (i.e., the transpose of A is equal to the negative of A).

Elementary Operation or Transformation of a Matrix:

Elementary row operations include:

Interchanging any two rows.
Multiplying a row by a non-zero scalar.
Adding or subtracting a multiple of one row from another row.

Inverse of a Matrix by Elementary Operations:

You can find the inverse of a matrix using elementary row operations. If the matrix A is invertible, you can transform it into the identity matrix I through row operations on an augmented matrix [A | I], where I is the identity matrix of the same order as A. If this process is successful, the resulting matrix on the left will be I, and the matrix on the right will be the inverse of A.

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Class 12 Maths Chapter 3 Question Answer (Intext Questions and Exercise)

Matrices Class 12 Questions And Answers: Exercise 3.1

Question:1(i). In the matrix $A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$ , write:

The order of the matrix

Answer:

$A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$

(i) The order of the matrix = number of row $\times$ number of columns $= 3\times 4$ .

Question 1(ii). In the matrix $A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$ , write:

The number of elements

Answer:

$A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(ii) The number of elements $3\times 4=12$ .

Question 1(iii). In the matrix $A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$ , write:

Write the elements a ₁₃ , a ₂₁ , a ₃₃ , a ₂₄ , a ₂₃

Answer:

$A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(iii) An element $a_{ij}$ implies the element in raw number i and column number j.

$a_1_3= 19$ $a_2_1= 35$

$a_3_3= -5$ $a_2_4= 12$

$a_2_3= \frac{5}{2}$

Question 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Answer:

A matrix has 24 elements.

The possible orders are :

$1\times 24,24\times 1,2\times 12,12\times 2,3\times 8,8\times 3,4\times 6 \, \, and\, \, 6\times 4$ .

If it has 13 elements, then possible orders are :

$1\times 13\, \, \, and \, \, \, \, 13\times 1$ .

Question 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

A matrix has 18 elements.

The possible orders are as given below

$1\times 18,18\times 1,2\times 9,9\times 2,3\times 6\, \, \, and\, \, \, \, 6\times 3$

If it has 5 elements, then possible orders are :

$1\times 5\, \, \, and \, \, \, \, 5\times 1$ .

Question 4(i). Construct a 2 × 2 matrix, $A = [a_{ij} ]$ whose elements are given by:

$a_{ij} = \frac{(i + j)^2}{2}$

Answer:

$A = [a_{ij} ]$

(i) $a_{ij} = \frac{(i + j)^2}{2}$

Each element of this matrix is calculated as follows

$a_1_1 = \frac{(1+1)^{2}}{2} =\frac{2^{2}}{2}=\frac{4}{2}=2$ $a_2_2 = \frac{(2+2)^{2}}{2} =\frac{4^{2}}{2}=\frac{16}{2}=8$

$a_1_2 = \frac{(1+2)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$ $a_2_1 = \frac{(2+1)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$

Matrix A is given by

$A = \begin{bmatrix} 2&4.5 \\4.5 & 8 \end{bmatrix}$

Question 4(ii). Construct a 2 × 2 matrix, $A = [a_{ij} ]$ , whose elements are given by:

$a_{ij} = \frac{i}{j}$

Answer:

A 2 × 2 matrix, $A = [a_{ij} ]$

(ii) $a_{ij} = \frac{i}{j}$

$a_1_1 = \frac{1}{1}=1$ $a_2_2 = \frac{2}{2}=1$

$a_1_2 = \frac{1}{2}$ $a_2_1 = \frac{2}{1}=2$

Hence, the matrix is

$A = \begin{bmatrix} 1& \frac{1}{2} \\ 2 & 1 \end{bmatrix}$

Question 4(iii). Construct a 2 × 2 matrix, $A = [a_{ij} ]$ , whose elements are given by:

$a_{ij} = \frac{(i+2j)^2}{2}$

Answer:

(iii)

$a_{ij} = \frac{(i+2j)^2}{2}$

$a_1_1 = \frac{(1+(2\times 1))^{2}}{2}= \frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$ $a_2_2 = \frac{(2+(2\times 2))^{2}}{2}= \frac{(2+4)^{2}}{2}=\frac{6^{2}}{2}=\frac{36}{2}=18$

$a_2_1 = \frac{(2+(2\times 1))^{2}}{2}= \frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}=8$ $a_1_2 = \frac{(1+(2\times 2))^{2}}{2}= \frac{(1+4)^{2}}{2}=\frac{5^{2}}{2}=\frac{25}{2}$

Hence, the matrix is given by

$A = \begin{bmatrix} \frac{9}{2}& \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

Question 5(i). Construct a 3 × 4 matrix, whose elements are given by:

$a_{ij} = \frac{1}{2}|-3i + j|$

Answer:

(i)

$a_{ij} = \frac{1}{2}|-3i + j|$

$a_1_1 = \frac{\left | -3+1 \right |}{2}=\frac{2}{2}=1$ $a_1_2 = \frac{\left | (-3\times 1)+2 \right |}{2}=\frac{1}{2}$ $a_1_3 = \frac{\left | (-3\times 1)+3 \right |}{2}=0$

$a_2_1 = \frac{\left | (-3\times 2)+1 \right |}{2}=\frac{5}{2}$ $a_2_2 = \frac{\left | (-3\times 2)+2 \right |}{2}=\frac{4}{2}=2$ $a_2_3 = \frac{\left | (-3\times 2)+3 \right |}{2}=\frac{\left | -6+3 \right |}{2}=\frac{\left | -3 \right |}{2} =\frac{3}{2}$

$a_3_1 = \frac{\left | (-3\times 3)+1 \right |}{2}=\frac{8}{2}=4$ $a_3_2 = \frac{\left | (-3\times 3)+2 \right |}{2}=\frac{7}{2}$ $a_3_3 = \frac{\left | (-3\times 3)+3 \right |}{2}=\frac{\left | -9+3 \right |}{2}=\frac{\left | -6 \right |}{2} =\frac{6}{2}=3$

$a_1_4 = \frac{\left | (-3\times 1)+4 \right |}{2}=\frac{\left | -3+4 \right |}{2}=\frac{\left | 1 \right |}{2} =\frac{1}{2}$ $a_2_4 = \frac{\left | (-3\times 2)+4 \right |}{2}=\frac{\left | -6+4 \right |}{2}=\frac{\left | -2 \right |}{2} =\frac{2}{2}=1$

$a_3_4 = \frac{\left | (-3\times 3)+4 \right |}{2}=\frac{\left | -9+4 \right |}{2}=\frac{\left | -5 \right |}{2} =\frac{5}{2}$

Hence, the required matrix of the given order is

$A = \begin{bmatrix} 1& \frac{1}{2} & 0&\frac{1}{2} \\ \frac{5}{2} & 2&\frac{3}{2}&1 \\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}$

Question 5(ii) Construct a 3 × 4 matrix, whose elements are given by:

$a_{ij} = 2i - j$

Answer:

A 3 × 4 matrix,

(ii) $a_{ij} = 2i - j$

$a_1_1 = 2\times 1-1 =2-1=1$ $a_1_2 = 2\times 1-2 =2-2=0$ $a_1_3 = 2\times 1-3 =2-3=-1$

$a_2_1 = 2\times 2-1 =4-1=3$ $a_2_2= 2\times 2-2 =4-2=2$ $a_2_3 = 2\times 2-3 =4-3=1$ $a_3_1 = 2\times 3-1 =6-1=5$ $a_3_2 = 2\times 3-2 =6-2=4$ $a_3_3 = 2\times 3-3 =6-3=3$

$a_1_4 = 2\times 1-4 =2-4=-2$ $a_2_4= 2\times 2-4 =4-4=0$ $a_3_4= 2\times 3-4 =6-4=2$

Hence, the matrix is

$A = \begin{bmatrix} 1 & 0& -1& -2 \\ \ 3 & 2&1& 0 \\5&4&3&2\end{bmatrix}$

Question 6(i). Find the values of x, y and z from the following equations:

$\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}$

Answer:

(i) $\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x=1\, \, \, ,\, \, \, y=4\, \, \, \, and\, \, \, \, z=3$

Question 6(ii) Find the values of x, y and z from the following equations:

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

Answer:

(ii)

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x+y=6$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

$x=6-y$

$xy=8$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$

Solving equation (i) and (ii) ,

$(6-y)y =8$

$6y-y^{2}=8$

$y^{2}-6y+8=0$

solving this equation we get,

$y=4 \, \, and\, \, y=2$

Putting the values of y, we get

$x=2 \, \, and\, \, x=4$

And also equating the first element of the second raw

$5+z = 5$ , $z=0$

Hence,

$x=2,y=4,z=0\, \, \, \, \, and\, \, \, \, \, \, x=4,y=2,z=0$

Question 6(iii) Find the values of x, y and z from the following equations

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

Answer:

(iii)

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$x+y+z=9........(1)$

$x+z=5..............(2)$

$y+z=7..............(3)$

subtracting (2) from (1) we will get y=4

substituting the value of y in equation (3) we will get z=3

now substituting the value of z in equation (2) we will get x=2

therefore,

$x=2$ , $y=4$ and $z=3$

Question 7. Find the value of a, b, c and d from the equation:

$\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}$

Answer:

$\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$a-b=-1$ $.............................1$

$2a+c=5$ $.............................2$

$2a-b=0$ $.............................3$

$3c+d=13$ $.............................4$

Solving equation 1 and 3 , we get

$a=1 \, \, \, \, and \, \, \, \, b=2$

Putting the value of a in equation 2, we get

$c=3$

Putting the value of c in equation 4 , we get

$d=4$

Question 8. $A = [a_{ij}]_{m\times n}$ is a square matrix, if

(A) $m <n$

(B) $m >n$

(D) None of these

Answer:

A square matrix has the number of rows and columns equal.

Thus, for $A = [a_{ij}]_{m\times n}$ to be a square matrix m and n should be equal.

Option (c) is .

Question 9. Which of the given values of x and y make the following pair of matrices equal

$\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$ , $\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

(A) $x = \frac{-1}{3}, y = 7$

(B) Not possible to find

(D) $x = \frac{-1}{3}, y = \frac{-2}{3}$

Answer:

Given, $\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$ $=\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$3x+7=0\Rightarrow x=\frac{-7}{3}$

$y-2=5 \Rightarrow y=5+2=7$

$y+1=8\Rightarrow y=8-1=7$

$2-3x=4\Rightarrow 3x=2-4\Rightarrow 3x=-2\Rightarrow x=\frac{-2}{3}$

Here, the value of x is not unique, so option B is correct.

Question 10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27
(B) 18
(C) 81
(D) 512

Answer:

Total number of elements in a 3 × 3 matrix

$=3\times 3=9$

If each entry is 0 or 1 then for every entry there are 2 permutations.

The total permutations for 9 elements

$=2^{9}=512$

Thus, option (D) is correct.

Class 12 Maths Chapter 3 Question Answer: Exercise 3.2

Question 1(i) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A + B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(i) A + B

The addition of matrix can be done as follows

$A+B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $+ \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A+B = \begin{bmatrix} 2+1 &4+3 \\ 3+(-2) & 2+5 \end{bmatrix}$

$A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}$

Question 1(ii) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A - B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(ii) A - B

$A-B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A-B = \begin{bmatrix} 2-1 &4-3 \\ 3-(-2) & 2-5 \end{bmatrix}$

$A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}$

Question 1(iii) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

3A - C

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

(iii) 3A - C

First multiply each element of A with 3 and then subtract C

$3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6 &12 \\ 9 & 6 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6-(-2) &12-5 \\ 9-3 & 6-4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 8 &7 \\ 6 & 2 \end{bmatrix}$

Question 1(iv) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(iv) AB

$AB = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $\times \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$AB = \begin{bmatrix} 2\times 1+4\times -2 & \, \, \, 2\times 3+4\times 5 \\ 3\times 1+2\times -2 & \, \, \, 3\times 3+2 \times 5 \end{bmatrix}$

$AB = \begin{bmatrix} -6 &26 \\ -1 & 19 \end{bmatrix}$

Question 1(v) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

Answer:

The multiplication is performed as follows

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$BA = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ $\times \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$

$BA = \begin{bmatrix} 1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+2\times 5 \end{bmatrix}$

$BA = \begin{bmatrix} 11 &10 \\ 11& 2 \end{bmatrix}$

Question 2(i). Compute the following:

$\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

Answer:

(i) $\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

$= \begin{bmatrix} a+a &b+b \\ -b+b & a+a \end{bmatrix}$

$= \begin{bmatrix} 2a &2b \\ 0 & 2a \end{bmatrix}$

Question 2(ii). Compute the following:

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

Answer:

(ii) The addition operation can be performed as follows

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

$=\begin{bmatrix} a^2 + b^2+2ab& b^2+c^2+2bc\\ a^2 + c^2-2ac& a^2 + b^2-2ab \end{bmatrix}$

$=\begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

Question 2(iii). Compute the following:

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

Answer:

(iii) The addition of given three by three matrix is performed as follows

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

$=\begin{bmatrix} -1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4 \end{bmatrix}$

$=\begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}$

Question 2(iv). Compute the following:

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

Answer:

(iv) the addition is done as follows

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

$=\begin{bmatrix} \cos^2+ \sin^2 x &\sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2x+ \sin^2 x \end{bmatrix}$ since $sin^2x+cos^2x=1$

$=\begin{bmatrix} 1 &1\\ 1 & 1 \end{bmatrix}$

Question 3(i). Compute the indicated products.

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

Answer:

(i) The multiplication is performed as follows

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a\times a+b\times b &a\times -b+b\times a \\ -b\times a+a\times b &-b\times -b+a\times a \end{bmatrix}$

$=\begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{bmatrix}$

Question 3(ii). Compute the indicated products.

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

Answer:

(ii) the multiplication can be performed as follows

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

$=\begin{bmatrix} 1\times 2 &1\times 3&1\times 4\\ 2\times 2&2\times 3&2\times 4\\3\times 2&3\times 3&3\times 4 \end{bmatrix}$

$=\begin{bmatrix} 2 &3& 4\\ 4&6&8\\6&9&12 \end{bmatrix}$

Question 3(iii). Compute the indicated products.

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

Answer:

(iii) The multiplication can be performed as follows

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 1\times 1+(-2)\times 2 & 1\times 2+(-2)\times 3&1\times 3+(-2)\times 1\\ 2\times 1+3\times 2 & 2\times 2+3\times 3&2\times 3+3\times 1 \end{bmatrix}$

Question 3(iv). Compute the indicated products.

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

Answer:

(iv) The multiplication is performed as follows

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix}\times \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+3\times 0+4\times 3 \, \, & 2\times (-3)+3\times 2+4\times 0 \, \, & 2\times 5+3\times 4+4\times 5 \\ 3\times 1+4\times 0+5\times 3 \, \, & 3\times (-3)+4\times 2+5\times 0 & 3\times 5+4\times 4+5\times 5 \\ 4\times 1+5\times 0+6\times 3 \, \, & 4\times (-3)+5\times 2+6\times 0\, \, & 4\times 5+5\times 4+6\times 5 \end{bmatrix}$

$= \begin{bmatrix} 14 & 0 & 42\\ 18 & -1 & 56\\ 22 & -2 & 70 \end{bmatrix}$

Question 3(v). Compute the indicated products.

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

Answer:

(v) The product can be computed as follows

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+1\times (-1) &2\times 0+1\times (2) & 2\times 1+1\times (1) \\ 3\times 1+2\times (-1) & 3\times 0+2\times (2) &3\times 1+2\times (1) \\ (-1)\times 1+1\times (-1) & (-1)\times 0+1\times (2) & (-1)\times 1+1\times (1) \end{bmatrix}$

$=\begin{bmatrix} 1 &2&3 \\ 1 & 4&5\\ -2 & 2&0 \end{bmatrix}$

Question 3(vi). Compute the indicated products.

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

Answer:

(vi) The given product can be computed as follows

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}$

$=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Question 4. If $A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ , $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$ , then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

$A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ , $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1+3 &2+(-1) &-3+2 \\ 5+4 &0+2 &2+5 \\ 1+2 & -1+0 &1+3 \end{bmatrix}$

$A+B = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}$

$B-C = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ $-\begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$B-C = \begin{bmatrix} 3-4 &-1-1 &2-2 \\ 4-0 &2-3 &5-2 \\ 2-1 & 0-(-2) &3-3 \end{bmatrix}$

$B-C = \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}$

Now, to prove A + (B - C) = (A + B) - C

$L.H.S\, \, :\, A+(B-C)$

$A+(B-C)=\begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}$ (Puting value of $B-C$ from above)

$A+(B-C)=\begin{bmatrix} 1-1 &2-2 &-3+0 \\ 5+4 &0+(-1) &2+3 \\ 1+1 & -1+2 &1+0 \end{bmatrix}$

$A+(B-C)=\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

$R.H.S\, \, :\, (A+B)-C$

$(A+B)-C = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}$ $- \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 4-4 &1-1 &-1-2 \\ 9-0 &2-3 &7-2 \\ 3-1 & -1-(-2) &4-3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Hence, we can see L.H.S = R.H.S = $\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Question 5. If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$ , then compute 3A - 5B

Answer:

$A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = 3\times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ $-5\times \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$ $- \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$

$3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$3A-5B = 0$

Question 6. Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$ .

Answer:

The simplification is explained in the following step

$\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta & \sin\theta \cos\theta \\ -\sin\theta \cos\theta & \cos^{2}\theta \end{bmatrix} +\begin{bmatrix} \sin^{2}\theta & - \sin\theta \cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta & \sin\theta \cos\theta - \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \sin\theta \cos\theta & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix}$

$= \begin{bmatrix} 1&0 \\ 0 & 1\end{bmatrix} =I$

the final answer is an identity matrix of order 2

Question 7(i). Find X and Y, if

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

Answer:

(i) The given matrices are

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}.............................1$

$X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}.............................2$

Adding equation 1 and 2, we get

$2 X = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ $+ \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$2 X = \begin{bmatrix} 7+3 &0+0 \\ 2+0 &5+3 \end{bmatrix}$

$2 X = \begin{bmatrix} 10 &0 \\ 2 &8 \end{bmatrix}$

$X = \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

Putting the value of X in equation 1, we get

$\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$ $+Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$

$Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} -$ $\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

$Y = \begin{bmatrix} 7-5 &0-0 \\ 2-1 &5-4 \end{bmatrix}$

$Y = \begin{bmatrix} 2 &0 \\ 1 &1 \end{bmatrix}$

Question 7(ii). Find X and Y, if

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

Answer:

(ii) $2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}..........................1$

$3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}......................2$

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

$3(2X + 3Y)-2(3X+2Y) = 3 \times \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ $- \, \, \, 2\times \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$6X + 9Y-6X-4Y= \begin{bmatrix} 6 &9 \\ 12 & 0 \end{bmatrix}$ $- \begin{bmatrix} 4 &-4 \\ -2 & 10 \end{bmatrix}$

$9Y-4Y= \begin{bmatrix} 6-4 &9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}$

$5Y= \begin{bmatrix} 2 &13 \\ 14 & -10 \end{bmatrix}$

$Y= \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$

Putting value of Y in equation 1 , we get

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + 3 \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix}$

$2X = \begin{bmatrix} 2-\frac{6}{5} &3-\frac{39}{5} \\ 4-\frac{42}{5} & 0 -(-6)\end{bmatrix}$

$2X = \begin{bmatrix} \frac{4}{5} &-\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix}$

$X = \begin{bmatrix} \frac{2}{5} &-\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix}$

Question 8. Find X, if $Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$ and $2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Answer:

$Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Substituting the value of Y in the above equation

$2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

$2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$

$2X = \begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}$

$X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$

Question 9. Find x and y, if $2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Answer:

$2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2+y & 6+0\\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2+y & 6\\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Now equating LHS and RHS we can write the following equations

$2+y=5$ $2x+2=8$

$y=5-2$ $2x=8-2$

$y=3$ $2x=6$

$x=3$

Question 10. Solve the equation for x, y, z and t, if $2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Answer:

$2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Multiplying with constant terms and rearranging we can rewrite the matrix as

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - 3\begin{bmatrix} 1& -1\\ 0 & 2 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - \begin{bmatrix} 3& -3\\ 0 & 6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9-3 &15-(-3)\\ 12-0 & 18-6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 6 &18\\ 12 & 12 \end{bmatrix}$

Dividing by 2 on both sides

$\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}$

$x=3,y=6,z=9\, \, and\, \, t=6$

Question 11. If $x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$ , find the values of x and y.

Answer:

$x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix} -y\\y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

Adding both the matrix in LHS and rewriting

$\begin{bmatrix}2x-y\\3x+y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$2x-y=10........................1$

$3x+y=5........................2$

Adding equation 1 and 2, we get

$5x=15$

$x=3$

Put the value of x in equation 2, we have

$3x+y=5$

$3\times 3+y=5$

$9+y=5$

$y=5-9$

$y=-4$

Question 12. Given $3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}$ , find the values of x, y, z and w.

Answer:

$3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}$

$\begin{bmatrix}3x &3 y \\3 z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$

If two matrices are equal than corresponding elements are also equal.

Thus, we have

$3x=x+4$

$3x-x=4$

$2x=4$

$x=2$

$3y=6+x+y$

Put the value of x

$3y-y=6+2$

$2y=8$

$y=4$

$3w=2w+3$

$3w-2w=3$

$w=3$

$3z=-1+z+w$

$3z-z=-1+3$

$2z=2$

$z=1$

Hence, we have $x=2,y=4,z=1\, \, and\, \, w=3.$

Question 13. If $F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$ , show that $F(x) F(y) = F(x + y)$ .

Answer:

$F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$

To prove : $F(x) F(y) = F(x + y)$

$R.H.S : F(x + y)$

$F(x+y) = \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

$L.H.S : F(x) F(y)$

$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}\times \begin{bmatrix} \cos y & -\sin y& 0\\\sin y &\cos y & 0 \\ 0 &0&1\end{bmatrix}$

$F(x)F(y) = \begin{bmatrix} \cos x \cos y- \sin x\sin y+0 & -\cos x \sin y-\sin x\cos y+0& 0+0+0\\\ sin x\cos y+\cos x \sin y+0 & - \sin x\sin y+\cos x \cos y+0 &0+0+0 \\ 0+0+0 &0+0+0&0+0+1\end{bmatrix}$

$F(x) F(y)= \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

Hence, we have L.H.S. = R.H.S i.e. $F(x) F(y) = F(x + y)$ .

Question 14(i). Show that

$\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

Answer:

To prove:

$\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

$L.H.S : \begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}5\times 2+(-1)\times 3 &5\times 1+(-1)\times 4\\6\times 2+7\times 3&6\times 1+7\times 4 \end{bmatrix}$

$= \begin{bmatrix}7 &1\\33&34 \end{bmatrix}$

$R.H.S : \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

$= \begin{bmatrix} 2\times 5+1\times 6 & 2\times (-1)+1\times 7\\ 3\times 5+4\times 6 & 3\times (-1)+4\times 7 \end{bmatrix}$

$= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}$

Hence, the right-hand side not equal to the left-hand side, that is

Question 14(ii). Show that

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

Answer:

To prove the following multiplication of three by three matrices are not equal

$L.H.S: \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}1\times(-1)+2\times 0+3\times 2 \, \, \, & 1\times(1)+2\times (-1)+3\times 3\, \, \, &1\times(0)+2\times 1+3\times 4\\0\times(-1)+1\times 0+0\times 2\, \, \, &0\times(1)+1\times (-1)+0\times 3\, \, \, &0\times(0)+1\times 1+0\times 4\\1\times(-1)+1\times 0+0\times 2\, \, \, &1\times(1)+1\times (-1)+0\times 3\, \, \, &1\times(0)+1\times 1+0\times 4 \end{bmatrix}$

$= \begin{bmatrix}5& 8&14\\0&-1&1\\-1&0&1\end{bmatrix}$

$R.H.S : \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

$= \begin{bmatrix}-1\times(1)+1\times 0+0\times 1 \, \, \, & -1\times(2)+1\times (1)+0\times 1\, \, \, &-1\times(3)+1\times 0+0\times 0\\0\times(1)+-(1)\times 0+1\times 1\, \, \, &0\times(2)+(-1)\times (1)+1\times 1\, \, \, &0\times(3)+(-1)\times 0+1\times 0\\2\times(1)+3\times 0+4\times 1\, \, \, &2\times(2)+3\times (1)+4\times 1\, \, \, &2\times(3)+3\times 0+4\times 0 \end{bmatrix}$

$= \begin{bmatrix}-1& -1&-3\\1&0&0\\6&11&6\end{bmatrix}$

Hence, $L.H.S \neq R.H.S$ i.e. $\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$ .

Question 15. Find $A^2 -5A + 6I$ , if

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

First, we will find ou the value of the square of matrix A

$A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times -1 & 2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 1& 2\times 0+1\times 1+3\times -1 &2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times -1 & 1\times 1+(-1)\times 3+0\times 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^2 -5A + 6I$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $-5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$ $+6 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $- \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 &15 \\ 5 & -5 & 0 \end{bmatrix}$ $+\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 &0 \\ 0 & 0 & 6 \end{bmatrix}$

$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0\\ 9-10+0 & -2-5+6 &5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}$

$= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}$

Question 16. If $A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$ prove that $A^3 - 6A^2 + 7A + 2I = 0$ .

Answer:

$A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

$A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$ $\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9 \end{bmatrix}$

$A^{2} = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$

$A^{3}=A^{2}\times A$

$A^{2}\times A = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$ $\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{3} = \begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39 \end{bmatrix}$

$A^{3} = \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^3 - 6A^2 + 7A + 2I = 0$

L.H.S :

$\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$ $- 6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$ $+7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$ $+2 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$ $- \begin{bmatrix}30&0&48\\12&24&30\\48&0&78 \end{bmatrix}$ $+ \begin{bmatrix}7&0&14\\0&14&7\\14&0&21 \end{bmatrix}$ $+ \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 &0 \\ 0 & 0 & 2 \end{bmatrix}$

$=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2 \end{bmatrix}$

$=\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{bmatrix}=0$

Hence, L.H.S = R.H.S

i.e. $A^3 - 6A^2 + 7A + 2I = 0$ .

Question 17. If $A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ and $I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$ , find k so that $A^{2} = kA - 2I$ .

Answer:

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $\times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$A^{2} = \begin{bmatrix}9-8 &-6+4\\12-8&-8+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}$

$A^{2} = kA - 2I$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$ $k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$ $2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$ $k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}+$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$ $=k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$\begin{bmatrix}1+2 &-2+0\\4+0&-4+2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

$\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

We have, $3=3k$

$k=\frac{3}{3}=1$

Hence, the value of k is 1.

Question 18. If $A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$ and I is the identity matrix of order 2, show that $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

To prove : $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

L.H.S : $I+A$

$I+A = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$ $+ \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1+0&0-\tan\frac{\alpha}{2}\\0+\tan\frac{\alpha}{2}&1+ 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

R.H.S : $(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $= (\begin{bmatrix}1 &0\\0&1 \end{bmatrix}-$ $\begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix})$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $=\begin{bmatrix} 1-0&0-(-\tan\frac{\alpha}{2})\\0-\tan\frac{\alpha}{2}&1- 0\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $=\begin{bmatrix} 1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$=\begin{bmatrix} \cos\alpha + \sin \alpha\tan\frac{\alpha}{2} &- \sin \alpha+ \cos \alpha \tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} \cos\alpha + \sin \alpha &\tan\frac{\alpha}{2} \sin\alpha + \cos \alpha \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}\tan\frac{\alpha}{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ (2\cos^{2} \frac{\alpha }{2} -1)\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} (2\cos^{2} \frac{\alpha }{2} -1) + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} &\tan\frac{\alpha}{2} 2\sin\frac{\alpha } {2} \ cos \frac{\alpha }{2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin^{2}\frac{\alpha }{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} -\tan\frac{\alpha}{2}\\-2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+\tan\frac{\alpha}{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} & 2\sin^{2}\frac{\alpha } {2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$= \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

Hence, we can see L.H.S = R.H.S

i.e. $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ .

Question 19(i). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix}x &(30000-x) \end{bmatrix}$ $\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix}$ $=1800$ (simple interest for 1 year $=\frac{pricipal\times rate}{100}$ )

$\frac{5}{100}x+\frac{7}{100}(30000-x) = 1800$

$5x+210000-7x=180000$

$210000-180000=7x-5x$

$30000=2x$

$x=15000$

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix}x &(30000-x) \end{bmatrix}$ $\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix}$ $=2000$ (simple interest for 1 year $=\frac{pricipal\times rate}{100}$ )

$\frac{5}{100}x+\frac{7}{100}(30000-x) = 2000$

$5x+210000-7x=200000$

$210000-200000=7x-5x$

$10000=2x$

$x=5000$

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount the bookshop will receive from selling all the books:

$12$ $\begin{bmatrix}10 &8&10 \end{bmatrix}$ $\begin{bmatrix}80\\60\\40 \end{bmatrix}$

$=12(10\times 80+8\times 60+10\times 40)$

$= 12(800+480+ 400)$

$= 12(1680)$

$=20160$

The total amount the bookshop will receive from selling all the books is 20160.

Question 21 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k , respectively. Choose the correct answer in Exercises 21 and 22.

Q21. The restriction on n, k and p so that PY + WY will be defined are:
(A) $k = 3, p = n$

(B) k is arbitrary, $p = 2$

(D) $k = 2, p = 3$

Answer:

P and Y are of order $p*k$ and $3*k$ respectivly.

$\therefore$ PY will be defined only if k=3, i.e. order of PY is $p*k$ .

W and Y are of order $n*3$ and $3*k$ respectivly.

$\therefore$ WY is defined because the number of columns of W is equal to the number of rows of Y which is 3, i.e. the order of WY is $n*k$

Matrices PY and WY can only be added if they both have same order i.e = $n*k$ implies p=n.

Thus, $k = 3, p = n$ are restrictions on n, k and p so that PY + WY will be defined.

Option (A) is correct.

Question 22 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k ,
respectively. Choose the correct answer in Exercises 21 and 22. If n = p , then the order of the matrix $7X - 5Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

X has of order $2*n$ .

$\therefore$ 7X also has of order $2*n$ .

Z has of order $2*p$ .

$\therefore$ 5Z also has of order $2*p$ .

Mtarices 7X and 5Z can only be subtracted if they both have same order i.e $2*n$ = $2*p$ and it is given that p=n.

We can say that both matrices have order of $2*n$ .

Thus, order of $7X - 5Z$ is $2*n$ .

Option (B) is correct.

Class 12 Maths Chapter 3 Question Answer: Exercise 3.3

Question 1(i). Find the transpose of each of the following matrices:

$\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

The transpose of the given matrix is

$A^{T}=\begin{bmatrix} 5& \frac{1}{2} &-1 \end{bmatrix}$

Question 1(ii). Find the transpose of each of the following matrices:

$\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

interchanging the rows and columns of the matrix A we get

$A^{T}=\begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix}$

Question 1(iii) Find the transpose of each of the following matrices:

$\begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Transpose is obtained by interchanging the rows and columns of matrix

$A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}$

Question 2(i). If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$ , then verify

$(A + B)' = A' + B'$

Answer:

$A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$(A + B)' = A' + B'$

L.H.S : $(A + B)'$

$A+B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$A+B = \begin{bmatrix} -1+(-4) & 2+1 & 3+(-5)\\ 5+1 &7+2 &9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix}$

$A+B = \begin{bmatrix} -5 & 3 & -2\\ 6 &9 &9 \\ -1 & 4 & 2 \end{bmatrix}$

$(A+B)' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}$

R.H.S : $A' + B'$

$A'+B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}$

$A'+B' = \begin{bmatrix} -1+(-4) & 5+1 & -2+1\\ 2+1 &7+2 &1+3 \\ 3+(-5) & 9+0 & 1+1 \end{bmatrix}$

$A'+B' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}$

Thus we find that the LHS is equal to RHS and hence verified.

Question 2(ii). If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$ , then verify

$(A - B)' = A' - B'$

Answer:

$A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$(A - B)' = A' - B'$

L.H.S : $(A - B)'$

$A-B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ $- \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$A-B = \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5)\\ 5-1 &7-2 &9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix}$

$A-B = \begin{bmatrix} 3 & 1 & 8\\ 4 &5 &9 \\ -3 & -2& 0 \end{bmatrix}$

$(A-B)' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}$

R.H.S : $A' - B'$

$A'-B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix}$ $- \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}$

$A'-B' = \begin{bmatrix} -1-(-4) & 5-1 & -2-1\\ 2-1 &7-2 &1-3 \\ 3-(-5) & 9-0 & 1-1 \end{bmatrix}$

$A'-B' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}$

Hence, L.H.S = R.H.S. so verified that

$(A - B)' = A' - B'$ .

Question 3(i). If $A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$ , then verify

$(A + B)' = A' + B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A + B)' = A' + B'$

$L.H.S : (A + B)' =$

$A+B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 3+(-1) & -1+(-1)&0+1\\ 4+1 &2+2 & 1+3 \end{bmatrix}$

$A+B = \begin{bmatrix} 2 & -2&1\\ 5 &4 & 4 \end{bmatrix}$

$\therefore \, \, \, (A+B)' = \begin{bmatrix} 2 & 5\\ 1 &4\\1 & 4 \end{bmatrix}$

R.H.S: $A' + B'$

$A'+B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'+B' = \begin{bmatrix} 2 & 5\\ 1 &4 \\ 1 & 4 \end{bmatrix}$

Hence, L.H.S = R.H.S i.e. $(A + B)' = A' + B'$ .

Question 3(ii). If $A = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$ , then verify

$(A - B)' = A' - B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A - B)' = A' - B'$

$L.H.S : (A - B)' =$

$A-B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$ $- \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A-B = \begin{bmatrix} 3-(-1) & -1-(2)&0-1\\ 4-1 &2-2 & 1-3 \end{bmatrix}$

$A-B = \begin{bmatrix} 4 & -3&-1\\ 3 &0 & -2 \end{bmatrix}$

$\therefore \, \, \, (A-B)' = \begin{bmatrix} 4 & 3\\ -3 &0\\-1 & -2 \end{bmatrix}$

R.H.S: $A' - B'$

$A'-B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $- \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'-B' = \begin{bmatrix} 4 & 3\\ -3 &0 \\ -1 & -2 \end{bmatrix}$

Hence, L.H.S = R.H.S i.e. $(A - B)' = A' - B'$ .

Question 4. If $A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$ and $B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$ , then find $(A + 2B)'$

Answer:

$B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$

$A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$

$(A + 2B)'$ :

$A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$ $+2 \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$

$A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$ $+ \begin{bmatrix} -2 & 0\\ 2 & 4 \end{bmatrix}$

$A+2B = \begin{bmatrix} -2+(-2) & 1+0\\ 3+2 & 2+4 \end{bmatrix}$

$A+2B = \begin{bmatrix} -4 & 1\\ 5 & 6 \end{bmatrix}$

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

$(A+2B)' = \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}$

Question 5(i) For the matrices A and B, verify that $(AB)' = B'A'$ , where

$A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$ , $B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$ , $B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}$

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$ $\begin{bmatrix} -1& 2 &1 \end{bmatrix}$

$AB = \begin{bmatrix} -1&2&1\\4&-8&-4 \\-3 &6&3\end{bmatrix}$

$(AB)' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1 &-4&3\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$

$A' = \begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$ $\begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1&-4&3 \end{bmatrix}$

Hence, L.H.S =R.H.S

so it is verified that $(AB)' = B'A'$ .

Question 5(ii) For the matrices A and B, verify that $(AB)' = B'A'$ , where

$A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}$

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 0\\1 \\2 \end{bmatrix}$ $\begin{bmatrix} 1& 5 &7 \end{bmatrix}$

$AB = \begin{bmatrix} 0&0&0\\1&5&7 \\2 &10&14\end{bmatrix}$

$(AB)' = \begin{bmatrix} 0&1&2\\0&5&10 \\0 &7&14\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$

$A' = \begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$ $\begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 0&1&2\\0&5&10 \\0&7&14 \end{bmatrix}$

Heence, L.H.S =R.H.S i.e. $(AB)' = B'A'$ .

Question 6(i). If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ , then verify that $A'A =I$

Answer:

$A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

By interchanging rows and columns we get transpose of A

$A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S : $A'A$

$A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ $\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

Question 6(ii). If $A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$ , then verify that $A'A = I$

Answer:

$A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

By interchanging columns and rows of the matrix A we get the transpose of A

$A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S : $A'A$

$A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$ $\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

Question 7(i). Show that the matrix $A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

the transpose of A is

$A' = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

Since, $A' = A$ so given matrix is a symmetric matrix.

Question 7(ii) Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$ is a skew-symmetric matrix.

Answer:

$A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

The transpose of A is

$A' = \begin{bmatrix} 0 & -1 & 1\\ 1 & 0 &-1 \\- 1 & 1 &0 \end{bmatrix}$

$A' =- \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

$A' =- A$

Since, $A' =- A$ so given matrix is a skew-symmetric matrix.

Question 8(i). For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ , verify that

$(A + A')$ is a symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $+ \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1+1 & 5+6\\ 6+5 & 7+7 \end{bmatrix}$

$A + A'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

$(A + A')'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

We have $A+A'=(A + A')'$

Hence , $(A + A')$ is a symmetric matrix.

Question 8(ii) For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ , verify that

$(A - A')$ is a skew symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $- \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1-1 & 5-6\\ 6-5 & 7-7 \end{bmatrix}$

$A - A'= \begin{bmatrix}0 & -1\\ 1& 0 \end{bmatrix}$

$(A - A')'= \begin{bmatrix}0 & 1\\ -1& 0 \end{bmatrix}=-(A-A')$

We have $A-A'=-(A - A')'$

Hence , $(A - A')$ is a skew-symmetric matrix.

Question 9. Find $\frac{1}{2}(A+A')$ and $\frac{1}{2}(A-A')$ , when $A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

the transpose of the matrix is obtained by interchanging rows and columns

$A' = \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$ $+\begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})$

$\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0+0 & a+(-a) & b+(-b)\\ -a+a & 0+0 & c+(-c)\\ -b+b & -c+c & 0+0 \end{bmatrix})$

$\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = 0$

$\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$ $- \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})$

$\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0-0 & a-(-a) & b-(-b)\\ -a-a & 0-0 & c-(-c)\\ -b-b & -c-c & 0-0 \end{bmatrix})$

$\frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 2a &2 b\\ -2a & 0 & 2c\\ -2b & -2c & 0 \end{bmatrix}$

$\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

Question 10(i). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

$\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$ $+\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}$

Let

$B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}$ $=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}$

$B'=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$ $-\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$ $= \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$

$C'= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew symmetric matrix.

Represent A as sum of B and C.

$B+C = \begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}$ $+ \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$ $= \begin{bmatrix} 3 & 5\\ 1 & -1\end{bmatrix}=A$

Question:10(ii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

$\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$ $+ \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}$

Let

$B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}$ $= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$B'= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$ $- \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 0&0\\ 0 & 0&0 \\0&0&0\end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$ $=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C'=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$ $+\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$ $= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=A$

Question 10(iii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

$\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$ $+\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 1 & -5\\ 1& -4 & -4\\ -5 & -4 & 4 \end{bmatrix}$

Let

$B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 1 & -5\\ 1 & -4 & -4\\ -5 & -4 & 4 \end{bmatrix}$ $= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}$

$B'= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$ $-\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 5&3\\ -5 & 0&6 \\-3&-6&0\end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 5&3\\ -5&0 & 6\\-3&-6&0 \end{bmatrix}$ $=\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix}$

$C'=\begin{bmatrix} 0 &- \frac{5}{2}&-\frac{3}{2}\\ \frac{5}{2}&0 &- 3\\\frac{3}{2}&3&0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}$ $+\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix}$ $=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}=A$

Question 10(iv). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

$\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

Answer:

$A =\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$ $+\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$

Let

$B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$ $=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$

$B'=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$ $-\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}$ $= \begin{bmatrix} 0 & 3\\ -3 & 0 \end{bmatrix}$

$C'= \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$ $- \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$ $= \begin{bmatrix} 1 & 5\\ -1 & 2\end{bmatrix}=A$

Question 11 Choose the correct answer in the Exercises 11 and 12.

If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Answer:

If A, B are symmetric matrices then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew symmetric.

Option A is correct.

Question 12 Choose the correct answer in the Exercises 11 and 12.

If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ and $A+A' =I$ , then the value of $\alpha$ is

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{3}$

(D) $\frac{3\pi}{2}$

Answer:

$A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$

$A' = \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$

$A+A' = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ $+ \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$ $= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$A+A' = \begin{bmatrix} 2\cos\alpha & 0\\ 0 & 2\cos\alpha \end{bmatrix}$ $= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$2 cos \alpha=1$

$cos \alpha=\frac{1}{2}$

$\alpha=\frac{\pi}{3}$

Option B is correct.

NCERT solutions for class 12 maths chapter 3 Matrices: Exercise 3.4

Question 1 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$

Answer:

Use the elementary transformation we can find the inverse as follows

$A=\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix}1&-1\\0&5 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{5}$

$\Rightarrow$ $\begin{bmatrix}1&-1\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A$

$R_1\rightarrow R_1+R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A$

$\therefore A^{-1}= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}$

Question 2 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2&1\\1&1\end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2&1\\1&1\end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2&1\\1&1\end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\1&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}A$

$A^{-1}= \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}$

Thus we have obtained the inverse of the given matrix through elementary transformation

Question 3 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

Using elementary transformations

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix}1&3\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-3R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$ .

Question 4 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix} 2 &3 \\ 1 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-3R_2$

$\Rightarrow$ $\begin{bmatrix} -1 &0 \\ 1 & 1 \end{bmatrix}$ $= \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\Rightarrow$ $\begin{bmatrix} -1 &0 \\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}7&-3\\5&-2\end{bmatrix}A$

$R_1\rightarrow -R_1$

$\Rightarrow$ $\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}-7&3\\5&-2\end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}-7&3\\5&-2\end{bmatrix}$

Question 5 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$

Answer:

$A =\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-3R_1$

$\Rightarrow$ $\begin{bmatrix} 2 & 1\\ 1 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-3&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$ $= \begin{bmatrix}4&-1\\-3&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}4&-1\\-7&2 \end{bmatrix}$ .

Thus the inverse of matrix A is obtained.

Question 6 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

Use the elementary transformation

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 2\\ 1 &3 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 2\\ 0 &1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-1&2 \end{bmatrix}A$

$R_1\rightarrow R_1-2R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$ $= \begin{bmatrix}3&-5\\-1&2 \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}3&-5\\-1&2 \end{bmatrix}$ .

Question 7 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix} 3 & 1\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-1&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}2&-1\\-1&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}2&-1\\-5&3 \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}2&-1\\-5&3 \end{bmatrix}$ .

Thus the inverse of matrix A is obtained using elementary transformation.

Question 8 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 &1 \\ 3 & 4 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-3R_1$

$\Rightarrow$ $\begin{bmatrix}1&1\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-3&4 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}A$

Thus using elementary transformation inverse of A is obtained as

$\therefore A^{-1}=$ $\begin{bmatrix}4&-5\\-3&4 \end{bmatrix}$ .

Question 9 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1& 3\\ 2 & 7 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix} 1& 3\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-2&3 \end{bmatrix}A$

$R_1\rightarrow R_1-3R_2$

$\Rightarrow$ $\begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}A$

Thus using elementary transformation the inverse of A is obtained as

$\therefore A^{-1}=$ $\begin{bmatrix}7&-10\\-2&3 \end{bmatrix}$ .

Question 10 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\Rightarrow$ $\begin{bmatrix} 3 & -1\\ -1 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\1&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1+2R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}$ $= \begin{bmatrix}3&2\\1&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix}$ $= \begin{bmatrix}3&2\\4&3 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{2}$

$\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}3&2\\2&\frac{3}{2} \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}$ .

Thus the inverse of A is obtained using elementary transformation.

Question 11 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-3R_1$

$\Rightarrow$ $\begin{bmatrix} -1 & 0\\ 1 & -2 \end{bmatrix}$ $= \begin{bmatrix}1&-3\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2+R_1$

$\Rightarrow$ $\begin{bmatrix} -1 & 0\\ 0 & -2 \end{bmatrix}$ $= \begin{bmatrix}1&-3\\1&-2 \end{bmatrix}A$

$R_1\rightarrow -R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 & -2 \end{bmatrix}$ $= \begin{bmatrix}-1&3\\1&-2 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{-2}$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix}A$

thus the inverse of matrix A is

$\therefore A^{-1}=$ $\begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix}$ .

Question 12 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{6}$

$\Rightarrow$ $\begin{bmatrix} 1& -\frac{1}{2}\\ -2 & 1 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2+2R_1$

$\Rightarrow$ $\begin{bmatrix} 1& -\frac{1}{2}\\ 0 & 0 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{6}&0\\\frac{1}{3}&1 \end{bmatrix}A$

Hence, we can see all the zeros in the second row of the matrix in L.H.S so $A^{-1}$ does not exist.

Question 13 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow 2R_2+R_1$

$\Rightarrow$ $\begin{bmatrix} 2 & -3\\ 0 &1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\1&2 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1+3R_2$

$\Rightarrow$ $\begin{bmatrix} 2 & 0\\ 0 &1 \end{bmatrix}$ $= \begin{bmatrix}4&6\\1&2 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{2}$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$ $= \begin{bmatrix}2&3\\1&2 \end{bmatrix}A$

so the inverse of matrix A is

$\therefore A^{-1}=$ $\begin{bmatrix}2&3\\1&2 \end{bmatrix}$ .

Question 14 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{2}$

$\Rightarrow$ $\begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&\frac{1}{2} \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&\frac{-1}{2}\\0&\frac{1}{2} \end{bmatrix}A$

Hence, we can see all upper values of matirix are zeros in L.H.S so $A^{-1}$ does not exists.

Question 15 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_3$

$\Rightarrow$ $\begin{bmatrix} -1 & -1 & 1\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0&-1\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow -R_1$

$\begin{bmatrix} 1 & 1 & -1\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$ $= \begin{bmatrix}-1&0&1\\0&1&0 \\0&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 1 & -1\\ 0& 0 & 5\\ 3 & -2 & 2 \end{bmatrix}$ $= \begin{bmatrix}-1&0&1\\2&1&-2 \\0&0&1 \end{bmatrix}A$

$R_3\rightarrow R_3-3R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 1 & -1\\ 0& 0 & 5\\ 0 & -5 & 5 \end{bmatrix}$ $= \begin{bmatrix}-1&0&1\\2&1&-2 \\3&0&-2 \end{bmatrix}A$

$R_2\leftrightarrow R_3$

$\Rightarrow$ $\begin{bmatrix} 1 & 1 & -1\\ 0& -5 & 5\\ 0 & 0 & 5 \end{bmatrix}$ $= \begin{bmatrix}-1&0&1\\3&0&-2 \\2&1&-2 \end{bmatrix}A$

$R_2\rightarrow \frac{-R_2}{5}$

$\Rightarrow$ $\begin{bmatrix} 1 & 1 & -1\\ 0& 1 & -1\\ 0 & 0 & 5 \end{bmatrix}$ $= \begin{bmatrix}-1&0&1\\\frac{-3}{5}&0&\frac{2}{5} \\2&1&-2 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & -1\\ 0 & 0 & 5 \end{bmatrix}$ $= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-3}{5}&0&\frac{2}{5} \\2&1&-2 \end{bmatrix}A$

$R_3\rightarrow \frac{R_3}{5}$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & -1\\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-3}{5}&0&\frac{2}{5} \\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}A$

$R_2\rightarrow R_2+R_3$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-1}{5}&\frac{1}{5}&0\\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}A$

Thos the Inverse of A is

$\therefore A^{-1}=$ . $\begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-1}{5}&\frac{1}{5}&0\\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}$ .

Question 16 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$ $= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_2\rightarrow R_2+3R_1$ and $R_3\rightarrow R_3-2R_1$

$\Rightarrow$ $\begin{bmatrix} 1 &3 & -2\\ 0& 9 &-11 \\ 0 & -1 & 4 \end{bmatrix}$ $= \begin{bmatrix}1&0&0\\3&1&0 \\-2&0&1 \end{bmatrix}A$

$R_1\rightarrow R_1+3R_3$ and $R_2\rightarrow R_2+8R_3$

$\Rightarrow$ $\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & -1 & 4 \end{bmatrix}$ $= \begin{bmatrix}-5&0&3\\-13&1&8 \\-2&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_3\rightarrow R_3+R_2$

$\Rightarrow$ $\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 25 \end{bmatrix}$ $= \begin{bmatrix}-5&0&3\\-13&1&8 \\-15&1&9 \end{bmatrix}A$

$R_3\rightarrow \frac{R_3}{25}$

$\Rightarrow$ $\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}-5&0&3\\-13&1&8 \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A$

$R_1\rightarrow R_1-10R_3$ and $R_2\rightarrow R_2-21R_3$

$\Rightarrow$ $\begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A$

Thus the inverse of three by three matrix A is

$\therefore A^{-1}=$ . $\begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}$ .

Question 17 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$ $= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{2}$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{2}&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-5R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 1 & 3 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\0&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_3\rightarrow R_3-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & \frac{1}{2} \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\\frac{5}{2}&-1&1 \end{bmatrix}A$

$R_3\rightarrow 2R_3$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\ 5&-2&2 \end{bmatrix}A$

$R_1\rightarrow R_1+\frac{R_3}{2}$ and $R_2\rightarrow R_2-\frac{5}{2}R_3$

$\Rightarrow$ $\begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$ $= \begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}A$

Thus the inverse of A is obtained as

$\therefore A^{-1}=$ . $\begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}$ .

Question:18 Matrices A and B will be inverse of each other only if

(A) $AB = BA$

(B) $AB = BA = 0$

(D) $AB = BA = I$

Answer:

We know that if A is a square matrix of order n and there is another matrix B of same order n, such that $AB=BA=I$ , then B is inverse of matrix A.

In this case, it is clear that A is inverse of B.

Hence, m atrices A and B will be inverse of each other only if $AB=BA=I$ .

Option D is correct .

NCERT solutions for class 12 maths chapter - 3 Matrices: Miscellaneous exercise

Question:1 Let $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$ , show that $(aI + bA)^n = a^n I + na^{n-1} bA$ , where I is the identity matrix of order 2 and $n \in N$ .

Answer:

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove : $(aI + bA)^n = a^n I + na^{n-1} bA$

For n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for n=1.

Let result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

(aI + bA)^k^+^1 = (aI + bA)^k^(aI + bA)

$= (a^k I + ka^{k-1} bA)$ $(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

(aI + bA)^k^+^1 $=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

(aI + bA)^k^+^1 $=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

Hence, the result is true for n=k+1.

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$ where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$ , $n \in N$ .

Question 2. If $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ then show that $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$ , $n\in N$ .

Answer:

Given :

$A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$

To prove:

$A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$

For n=1, we have

$A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix}$ $=\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix}$ $= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ $\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$ , $n\in N$ where $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ .

Question 3. If $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$ , then prove that $A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}$ , where n is any positive integer.

Answer:

Given :

$A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$

To prove:

$A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}$

For n=1, we have

$A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}$ $= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A$

Thus, result is true for n=1.

Now, take result is true for n=k,

$A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$ $\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

$=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}$

$=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}$

$=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}$

$=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}$ , where $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$ .

Question 4. If A and B are symmetric matrices, prove that $AB - BA$ is a skew symmetric matrix.

Answer:

If A, B are symmetric matrices then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew symmetric.

Question 5. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Let be a A is symmetric matrix , then $A'=A$

Consider, $(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by $A$

$=B'(AB)$

i.e. $(B'AB)'$ $=B'(AB)$

Thus, if A is a symmetric matrix than $B'(AB)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A'=-A$ .

$(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by - $A$ ,

$=B'(-AB)$

$= - B'AB$

i.e. $(B'AB)'$ $= - B'AB$ .

Thus, if A is a skew-symmetric matrix then $- B'AB$ is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

Question 6. Find the values of x , y , z if the matrix $A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$ satisfy the equation $A'A = I$

Answer:

$A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$

$A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$

$A'A = I$

$\begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$ $\begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$ $= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix}$ $= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} 2x^{2} & 0& 0\\ 0& 6y^{2} & 0\\ 0 & 0 &3z^{2}\end{bmatrix}$ $= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

Thus equating the terms elementwise

$2x^{2} = 1$ $6y^{2} = 1$ $3z^{2} = 1$

$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2}=\frac{1}{3}$

$x = \pm \frac{1}{\sqrt{2}}$ $y= \pm \frac{1}{\sqrt{6}}$ $z=\pm \frac{1}{\sqrt{3}}$

Question 7. For what values of x: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$ ?

Answer:

$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 0+4+4x \end{bmatrix} = O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, value of x is -1.

Question 8. If $A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$ , show that $A^2 -5A + 7I= 0$ .

Answer:

$A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 9-1 &3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$

$I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

To prove: $A^2 -5A + 7I= 0$

L.H.S : $A^2 -5A + 7I$

$= \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$ $-5 \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$ $+ 7 \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 8-15+7 &5-5+0 \\ -5+5+0& 3-10+7 \end{bmatrix}$

$=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S$

Hence, we proved that

$A^2 -5A + 7I= 0$ .

Question 9. Find x, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$ .

Answer:

$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x +0-2& 0-10+0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x -2& -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix}x (x -2)-40+(2x-8) \end{bmatrix} = 0$

$\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0$

$\therefore \, \, x ^{2}-48= 0$

$x ^{2}=48$

thus the value of x is

$x =\pm 4\sqrt{3}$

Question 10(a) A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 10000\times 2.50+2000\times 1.50+18000\times 1.00$

$= 25000+3000+18000$

$= 46000$

The total revenue in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$= 15000+30000+8000$

$= 53000$

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Question 10(b). A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 10000\times 2.00+2000\times 1.00+18000\times 0.50$

$= 20000+2000+9000$

$= 31000$

Total revenue in the market I is 46000 , gross profit in the market is $= 46000-31000$ $=Rs. 15000$

The total cost price in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 6000\times 2.0+20000\times 1.0+8000\times 0.50$

$= 12000+20000+4000$

$= 36000$

Total revenue in market II is 53000, gross profit in the market is $= 53000-36000= Rs. 17000$

Question 11. Find the matrix X so that $X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

Answer:

$X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

The matrix given on R.H.S is $2\times 3$ matrix and on LH.S is $2\times 3$ matrix.Therefore, X has to be $2\times 2$ matrix.

Let X be $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$

$\begin{bmatrix} a & c\\ b & d \end{bmatrix}$ $\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$\begin{bmatrix} a+4c & 2a+5c &3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$a+4c=-7$ $2a+5c=-8$ $3a+6c=-9$

$b+4d=2$ $2b+5d=4$ $3b+6d=6$

Taking, $a+4c=-7$

$a=-4c-7$

$2a+5c=-8$

$-8c-14+5c=-8$

$-3c=6$

$c=-2$

$a=-4\times -2-7$

$a=8-7=1$

$b+4d=2$

$b=-4d+2$

$2b+5d=4$

$\Rightarrow$ $-8d+4+5d=4$

$\Rightarrow -3d=0$

$\Rightarrow d=0$

$b=-4d+2$

$\Rightarrow b=-4\times 0+2=2$

Hence, we have $a=1, b=2,c=-2,d=0$

Matrix X is $\begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix}$ .

Question 12. If A and B are square matrices of the same order such that $AB = BA$ , then prove by induction that $AB^n = B^n A$ . Further, prove that $(AB)^n = A^n B^n$ for all $n \in N$ .

Answer:

A and B are square matrices of the same order such that $AB = BA$ ,

To prove : $AB^n = B^n A$ , $n \in N$

For n=1, we have $AB^1 = B^1 A$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $AB^k = B^k A$

Now, taking n=k+1 , we have $AB^{k+1} = AB^k .B$

$AB^{k+1} = (B^kA) .B$

$AB^{k+1} = (B^k) .AB$

$AB^{k+1} = (B^k) .BA$

$AB^{k+1} = (B^k.B) .A$

$AB^{k+1} = (B^k^+^1) .A$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ , $n \in N$ .

To prove: $(AB)^n = A^n B^n$

For n=1, we have $(AB)^1 = A^1 B^1$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $(AB)^k = A^k B^k$

Now, taking n=k+1 , we have $(AB)^{k+1} = (A B)^k.(AB)$

$(AB)^{k+1} = A^k B^k.(AB)$

$(AB)^{k+1} = A^{K}( B^kA)B$

$(AB)^{k+1} = A^{K}( AB^k)B$

$(AB)^{k+1} = (A^{K}A)(B^kB)$

$(AB)^{k+1} = (A^{k+1})(B^{k+1})$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ and $(AB)^n = A^n B^n$ for all $n \in N$ .

Question 13 Choose the correct answer in the following questions:

If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$

(A) $1 + \alpha^2 + \beta \gamma = 0$

(B) $1 - \alpha^2 + \beta \gamma = 0$

(D) $1 + \alpha^2 - \beta \gamma = 0$

Answer:

$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$

$A^2 = I$

$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ $\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

Thus we obtained that

$\alpha^{2} +\beta \gamma=1$

$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$

Option C is correct.

Question 14. If the matrix A is both symmetric and skew-symmetric, then

(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these

Answer:

If the matrix A is both symmetric and skew-symmetric, then

$A'=A$ and $A'=-A$

$A'=A'$

$\Rightarrow \, \, \, \, \, \, \, A=-A$

$\Rightarrow \, \, \, \, \, \, \, A+A=0$

$\Rightarrow \, \, \, \, \, \, \, 2A=0$

$\Rightarrow \, \, \, \, \, \, \, A=0$

Hence, A is a zero matrix.

Option B is correct.

Question 15. If A is square matrix such that $A^{2}=A$ , then $(I + A)^3 - 7 A$ is equal to

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

A is a square matrix such that $A^{2}=A$

$(I + A)^3 - 7 A$

$=I^{3}+A^{3}+3I^{2}A+3IA^{2}-7A$

$=I+A^{2}.A+3A+3A^{2}-7A$

$=I+A.A+3A+3A-7A$ (Replace $A^{2}$ by $A$ )

$=I+A^{2}+6A-7A$

$=I+A-A$

$=I$

Hence, we have $(I + A)^3 - 7 A=I$

Option C is correct.

If you are interested in Matrices Class 12 NCERT Solutions exercises then these are listed below.

Matrices Class 12 NCERT Solutions Exercise 3.1

Matrices Class 12 NCERT Solutions Exercise 3.2

Matrices Class 12 NCERT Solutions Exercise 3.3

Matrices Class 12 NCERT Solutions Exercise 3.4

Matrices Class 12 NCERT Solutions Miscellaneous Exercise

Matrices Class 12 NCERT solutions - Topics

3.1 Introduction

3.2 Matrix

3.2.1 Order of a matrix

3.3 Types of Matrices

3.3.1 Equality of matrices

3.4 Operations on Matrices

3.4.1 Addition of matrices

3.4.2 Multiplication of a matrix by a scalar

3.4.3 Properties of matrix addition

3.4.4 Properties of scalar multiplication of a matrix

3.4.5 Multiplication of matrices

3.4.6 Properties of multiplication of matrices

3.5. Transpose of a Matrix

3.5.1 Properties of the transpose of the matrices

3.6 Symmetric and Skew-Symmetric Matrices

3.7 Elementary Operation (Transformation) of a Matrix

3.8 Invertible Matrices

3.8.1 Inverse of a matrix by elementary operations

Topics of NCERT Class 12 Maths Chapter Matrices

The main topics covered in maths chapter 3 class 12 are:

Matrix

It is an ordered representation of numbers and functions, known as elements in a rectangular array. In this class 12 NCERT topics discuss concepts related to the order of a matrix, elements in raw, and columns of a matrix. there are good quality questions in matrix class 12 solutions.

Type of matrices

This ch 1 maths class 12 concerns different types of matrices like column matrix, raw matrix, square matrix, diagonal matrix, scalar matrix, identity matrix, zero matrix, etc. In this ch 3 maths class 12 also discuss comprehensively conditions of equality of matrix. To get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 3.

Operations on matrices

This ch 3 maths class 12 also includes concepts of addition of matrices, multiplication of a matrix by a scalar, negative matrix, difference of matrices. maths class 12 chapter 3 also contains properties of matrix addition that include commutative leas, associative laws, the existence of an additive identity, the existence of additive inverse. Properties of scalar multiplication of matrix elaborated in this chapter. You can refer to class 12 NCERT solutions for questions about these concepts.

multiplication of matrix

Properties of multiplication of matrix that includes associative law, distributive laws, the existence of multiplicative identity. to get command of these concepts you can go through the NCERT solution for class 12 maths chapter 3.

transpose of a matrix

properties of transpose of the matrix are discussed in maths class 12 chapter 3. matrix class 12 solutions include quality questions to understand the concepts.

ch 3 maths class 12 also discuss in detail the concepts of symmetric and skew symmetric matrix, elementary operations (transformation) of a matrix. for questions on these concepts, you can browse NCERT solutions for class 12 chapter 3.

Invertible matrix

In class 12 NCERT chapter you get an idea of the inverse of a matrix by elementary operation. class 12 NCERT solutions include problems related to these concepts

Topics enumerated in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 3

Also read,

NCERT Solutions for Class 12 - Subject Wise

NCERT solutions for class 12 maths - Chapter wise

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

How to use NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT Class 12 Maths solutions chapter 3 covers all the questions given in the textbook. Make sure you have gone through all the important concepts of NCERT class 12th maths Matricesbefore solving the questions.
Firstly try to solve the questions on your own and then check the NCERT class 12 maths chapter 3 solutions.
Along with the NCERT Solutions for Class 12 Maths Chapter 3 Matrices, solve the previous year questions as well.
NCERT Class 12 maths chapter 3 solutions pdf download will be made available soon. Till then you can save this webpage instead of NCERT class 12 maths chapter 3 pdf to practice questions offline.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. How are the NCERT solutions helpful in the board exam?

NCERT solutions are not only important when you are stuck while solving the problems but you will get to know how to answer in the board exam in order to get good marks in the board exam. these class 12 maths ch 3 question answer also make you understand new concepts or in depth understanding of topics and can solve you doubts. in this way ncert solutions becomes important so you can refer ncert textbooks, ncert exercises and ncert solutions.

2. What are the important topics in chapter matrices?

Matrices, order of a matrix, types of matrices, equality of matrices, operations like addition multiplication on matrices, symmetric and skew-symmetric matrices are the important topics in this chapter. these topics are very important because of use in other supplementary topics like 3d geometry.

3. Can using class 12 maths chapter 3 ncert solutions improve your performance on the board exam?

Yes, Class 12 Matrices solutions definitely improve the performance on the board exam. It's recommended that students begin by tackling simpler problems before moving on to more challenging ones. Class 12 matrices NCERT solutions will enable them to identify areas where they need improvement. Through repeated practice on their weaker concepts, students can strengthen their skills and perform well on the board exams. Additionally, short-cut tips are provided to assist students in finding easier ways to solve complex problems.

4. What is the weightage of the chapter matrices for CBSE board exam?

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination. students can list out topics according to respective weightage and channelise their energy according to priority of the topics.

5. What are the primary themes covered in NCERT Solutions for class 12 chapter 3 maths?

Mathematics often has simple chapters that can be enjoyable to study once understood, and matrices are one example. NCERT Class 12 maths matrices focus on the following main topics: matrices, types of matrices, operations on matrices, transpose of a matrix, symmetric and skew symmetric matrices, elementary operations on matrices, and invertible matrices. The material is presented in a straightforward manner to help students achieve good grades in the board exams, regardless of their intelligence. For ease, students can study matrices class 12 ncert pdf both online and offline.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

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Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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Muskan Dhalwal 17 August,2022

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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