NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Matrices
Given a matrix A such that $\mathrm{A}=\left[a_{ij}\right]$ is an $m \times n$ matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. In this exercise, you will get NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.3, which will include questions related to properties of the transpose of the matrices, symmetric and skew-symmetric matrices in Exercise 3.3, Class 12 Maths. Going through these solutions will help you to understand the concept clearly. These NCERT solutions are created by a subject matter expert at Careers360 to give a more systematic and proper approach for each question. You should try to solve these Class 12th maths chapter 3 exercise 3.3 of the NCERT on your own. You can take help from these solutions, which are prepared by experts who know how best to answer in board exams
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The overall CBSE class 12th Maths exam 2026 can be considered balanced and moderately difficult. Students with a clear conceptual understanding and regular practice of NCERT and exemplar problems would have found the paper manageable. The question paper was a fair mix of knowledge, application, and analytical questions, reflecting the ongoing shift towards competency-based assessment in school education.
Class 12 Maths Chapter 3 Exercise 3.3 Solutions: Download PDF
Matrices Exercise: 3.3
Question 1(i). Find the transpose of each of the following matrices:
$\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$
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
$A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}$
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.
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
$(A - B)' = A' - B'$.
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
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. $(A - B)' = A' - B'$.
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}$
$B'A' = \begin{bmatrix} 0&1&2\\0&5&10 \\0&7&14 \end{bmatrix}$
Hence, 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 the 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 :$A'A$
$A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$ $\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$
$A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}$
$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$
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.
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.
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}$$= \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$
$C'= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$
$C=-C'$
Thus, $\frac{1}{2}(A-A')$ is a skew symmetric matrix.
Represent A as the sum of B and C.
$B+C = \begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}$ $+ \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$ $= \begin{bmatrix} 3 & 5\\ 1 & -1\end{bmatrix}=A$
Question:10(ii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
$\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$
Answer:
$A=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$
$A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$
$A+A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$$+ \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}$$= \begin{bmatrix} 0 & 3\\ -3 & 0 \end{bmatrix}$
$C'= \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$
$C=-C'$
Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.
Represent A as the sum of B and C.
$B+C=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$ $- \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$ $= \begin{bmatrix} 1 & 5\\ -1 & 2\end{bmatrix}=A$
Question 11 Choose the correct answer in the Exercises 11 and 12.
If A, B are symmetric matrices of the same order, then AB – BA is a
(A) Skew-symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
Answer:
If A, B are symmetric matrices, then
$A'=A$ and $B' = B$
we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$
$=BA-AB$
$= -(AB-BA)$
Hence, we have $(AB-BA) = -(AB-BA)'$
Thus,( AB-BA)' is skew symmetric.
Option A is correct.
Question 12 Choose the correct answer in the Exercises 11 and 12.
If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ and $A+A' =I$, then the value of $\alpha$ is
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{3}$
(C) $\pi$
(D) $\frac{3\pi}{2}$
Answer:
$A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$
$A' = \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$
$A+A' = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$$+ \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.
Also Read,
Topics covered in Chapter 3: Matrices: Exercise 3.3
- Intoduction
- Transpose of a matrix: If $\mathrm{A}=\left[a_{ij}\right]$ is an $m \times n$ matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A.
- Properties of the transpose of the matrices:
- $\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}$,
- $(k \mathrm{~A})^{\prime}=k \mathrm{~A}^{\prime}$ (where $k$ is any constant)
- $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$
- $(\mathrm{A} \mathrm{B})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}$
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Download EBook- Symmetric Matrix: A square matrix $\mathrm{A}=\left[a_{i j}\right]$ is said to be symmetric if $\mathrm{A}^{\prime}=\mathrm{A}$, that is, $\left[a_{i j}\right]=\left[a_{j i}\right]$ for all possible values of $i$ and $j$.
- Skew Symmetric Matrix: A square matrix $\mathrm{A}=\left[a_{i j}\right]$ is said to be skew symmetric matrix if $\mathrm{A}^{\prime}=-\mathrm{A}$, that is $a_{j i}=-a_{i j}$ for all possible values of $i$ and $j$. Now, if we put $i=j$, we have $a_{i i}=-a_{i i}$. Therefore $2 a_{i i}=0$ or $a_{i i}=0$ for all $i$ 's.
- Theorem 1: For any square matrix A with real number entries, $\mathrm{A}+\mathrm{A}^{\prime}$ is a symmetric matrix and $\mathrm{A}-\mathrm{A}^{\prime}$ is a skew-symmetric matrix.
- Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.
Also, read,
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Frequently Asked Questions (FAQs)
Q: How these NCERT textbook solutions are helpful in board exam ?
A:
NCERT solutions will help you to solve NCERT problems when you are not able to solve them on your own. For more questions NCERT exemplar problems will be useful. For CBSE board exam NCERT syllabus will be useful for exam preparation. Practice class 12 ex 3.3 to command the concepts.
Q: What is the definition of order of a Matrix ?
A:
The order of matrix having m rows and n columns is m x n.
Q: What is symmetric matrix ?
A:
If the transpose of matrix A is equal to matrix A then matrix A is a symmetric matrix.
Q: What is skew-symmetric matrix ?
A:
If the transpose of matrix A is equal to the negative of matrix A then matrix A is a skew-symmetric matrix.
Q: What are the diagonal elements of skew symmetric matrix ?
A:
All the diagonal elements of a skew-symmetric matrix are zero.
Q: What is the transpose of A' ?
A:
(A')' = A
Hence the transpose of A' is matrix A.
Q: If A is symmetric matrix then A' ?
A:
If A is a symmetric matrix then A' = A.
Q: If A is symmetric matrix and k is a constant then (kA) ' ?
A:
If A is a symmetric matrix and k is a constant then (kA) ' = k (A)'
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