# NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Matrices

In this article, you will get NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 which consists of questions related to the transpose of a matrix. If a matrix A of order m x n, then by interchanging the rows and columns of A we obtain the transpose of matrix A. Also, you will get questions related to properties of the transpose of the matrices, symmetric and skew-symmetric matrices in the exercise 3.3 class 12 maths. Only going through these class 12 maths ch 3 ex 3.3 solutions won't help you to understand the concept clearly. You should try to solve these class 12th maths chapter 3 exercise 3.3 questions on your own. You can take help from these solutions which are prepared by experts who know how best to answer in board exams. You can check for NCERT solutions for class 12 maths.

Also, see

Matrices Exercise 3.1

Matrices Exercise 3.2

Matrices Exercise 3.4

Matrices Miscellaneous Exercise

## Matrices Exercise: 3.3

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

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

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

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:

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $(A + 2B)'$

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

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

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

$(A + 2B)'$ :

$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

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

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

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

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

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

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 $A'A =I$

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

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 $A'A = I$

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

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 :$A'A$

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

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

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

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

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

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.

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

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.

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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:

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

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

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

$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:

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

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

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

$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:

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

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

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

$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:

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

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

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

$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

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

(C) $\pi$

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

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

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

$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.

## More About NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3:-

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 consists of 12 descriptive questions including two multiple choice type questions. Exercise 3.3 Class 12 Maths questions are related to finding the transpose of the matrices and the application of properties of transpose. There are 4 solved examples given before the NCERT textbook questions in the. First, try to solve these examples which will help you to get conceptual clarity. It will also help you in solving NCERT book problems easily as most of the textbook problems are related to solved examples.

Also Read| Matrices Class 12 Maths Chapter Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3:-

• Class 12 Maths chapter 3 exercise 3.3 solutions are helpful when you find it difficult to solve NCERT syllabus problems on your own.
• These solutions are designed in a descriptive manner which will help you to get conceptual clarity.
• NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 are helpful for the students to secure good marks in the board exam.
• Exercise 3.3 Class 12 Maths solutions can be used for reference also.

Also see-

• NCERT solutions for Class 12 Maths chapter 3

• NCERT exemplar solutions Class 12 Maths chapter 3

## NCERT Solutions of cCass 12 Subject Wise

• NCERT solutions for Class 12 Maths

• NCERT solutions for Class 12 Physics

• NCERT solutions for Class 12 Chemistry

• NCERT solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

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