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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 03:45 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.3

NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 Matrices are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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 p roperties 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.

12th class Maths exercise 3.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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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}

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} \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.

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}

(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.

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.
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Key Features Of NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 3.3 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 3.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 3.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 3.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 3.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 3.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of cCass 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. How these NCERT textbook solutions are helpful in board exam ?

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.

2. What is the definition of order of a Matrix ?

The order of matrix having m rows and n columns is m x n. 

3. What is symmetric matrix ?

If the transpose of matrix A is equal to matrix A then matrix A is a symmetric matrix.

4. What is skew-symmetric matrix ?

If the transpose of matrix A is equal to the negative of matrix A then matrix A is a skew-symmetric matrix.

5. What are the diagonal elements of skew symmetric matrix ?

All the diagonal elements of a skew-symmetric matrix are zero.

6. What is the transpose of A' ?

(A')' = A

Hence the transpose of A' is matrix A.

7. If A is symmetric matrix then A' ?

If A is a symmetric matrix then A' = A.

8. If A is symmetric matrix and k is a constant then (kA) ' ?

If A is a symmetric matrix and k is a constant then (kA) ' = k (A)'

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer
Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
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