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
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)
- $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$
- $(\mathrm{A} \mathrm{B})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}$
JEE Main Highest Scoring Chapters & Topics
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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,
NCERT Solutions Subject Wise
These links lead to NCERT textbook solutions for other subjects. Students can check and analyse these well-structured solutions for a deeper understanding.
CBSE Class 12th Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters
Subject-wise NCERT Exemplar solutions
Students may visit these NCERT exemplar links for additional practice.
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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The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.
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Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.
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Hello Pruthvi,
Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.
The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.
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Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
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For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.
Popular CBSE Class 12th Questions
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0.02 |
Option 2)
3.125 × 10-2 |
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2.5 × 10-2 |
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