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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
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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 Questions (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 ?
Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

  1. Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
  2. Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
  3. Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
  4. Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

Read Complete Answer
Prachi Kumari 30 December,2024
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
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.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,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,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,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
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

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0.16\; J
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1.00\; J

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

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2.45×10−3 kg

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 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

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12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

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6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

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3.125 × 10-2
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1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

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increase two fold
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remain unchanged

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be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

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Mole fraction.
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twice that in 60 g carbon

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

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more than 3 but less than 6
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more than 6 but less than 9

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