# NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 (Ex 3.3)

## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 (Ex 3.3)

Solved NCERT Solutions of Ex 3.3 Class 12 Maths, are provided by Vedantu in an easy to understand way. Students appearing for the 12th Board exams generally find it difficult while revising the Mathematics chapters. To make the toughest concepts easy and understand the complex formulae, Vedantu NCERT Solutions are prepared by highly qualified teachers. Download the free PDF files and understand Exercise 3.3 Maths Class 12 from experts in an easy way. We provide solutions and study material that is prepared as per the recent NCERT guidelines.

### Some important points to learn before solving Exercise 3.3

• A matrix's transpose is obtained by swapping its rows and columns. A matrix is a rectangular array of numbers or functions that are arranged in rows and columns. This array of numbers is referred to as elements of a matrix.

• The transpose of a matrix is an operator in linear algebra that flips a matrix over its diagonal by changing the row and column indices of the matrix thereby, producing another matrix. Generally, the transpose of a matrix A is denoted by either A' or AT.

• A matrix's order represents the number of rows and columns of the given matrix. The horizontal lines of the elements are all referred to as the rows of the matrix, denoted by n, and the vertical lines of the elements are referred to as the columns of the matrix, denoted by m. They indicate the order of a matrix, which is denoted by n x m. The order of the given matrix's transpose is expressed as m x n.

• If we consider two matrices B & C, then the transpose of the sum (B + C)T will be the sum of transposes of the matrices B & C. This can be given as (B + C)T = BT+CT.

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## Access NCERT Solutions for Class 12 Maths Chapter 3- Matrix

### Exercise 3.3

1. Find the transpose of each of the following matrices:

\begin{align} & \mathrm{(i)}\left[ \begin{array}{*{35}{l}} \mathrm{5} \\ \mathrm{1/2} \\ \mathrm{-1} \\ \end{array} \right] \\ & \mathrm{(ii)}\left[ \begin{matrix} \mathrm{1} & \mathrm{-1} \\ \mathrm{2} & \mathrm{3} \\ \end{matrix} \right] \\ & \mathrm{(iii)}\left[ \begin{matrix} \mathrm{-1} & \mathrm{5} & \mathrm{6} \\ \sqrt{\mathrm{3}} & \mathrm{5} & \mathrm{6} \\ \mathrm{2} & \mathrm{3} & \mathrm{-1} \\ \end{matrix} \right] \\ \end{align}

Ans:

Find transpose

$(i)A=\left[ \begin{array}{*{35}{l}} 5 \\ 1/2 \\ -1 \\ \end{array} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} 5 & \frac{1}{2} & -1 \\ \end{matrix} \right]$ $(ii)A=\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} 1 & 2 \\ -1 & 3 \\ \end{matrix} \right]$ $(iii)A=\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \\ \end{matrix} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \\ \end{matrix} \right]$

2. If $\mathrm{A=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{3} \\ \mathrm{5} & \mathrm{7} & \mathrm{9} \\ \mathrm{-2} & \mathrm{1} & \mathrm{1} \\ \end{matrix} \right]$ and $\mathrm{B=}\left[ \begin{matrix} \mathrm{-4} & \mathrm{1} & \mathrm{-5} \\ \mathrm{1} & \mathrm{2} & \mathrm{0} \\ \mathrm{1} & \mathrm{3} & \mathrm{1} \\ \end{matrix} \right]$ , then verify that \begin{align} & \mathrm{(i)}\left( \mathrm{A+B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ +B }\!\!'\!\!\text{ } \\ & \mathrm{(ii)}\left( \mathrm{A-B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ -B }\!\!'\!\!\text{ } \\ \end{align}

Ans:

We have

\begin{align} & A'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]B'=\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right] \\ & (i)A+B=\left[ \begin{matrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \\ \end{matrix} \right]\Rightarrow \left( A+B \right)'=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right] \\ & A'+B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right] \\ \end{align} Hence, we have verified that $\left( A+B \right)'=A'+B'$ \begin{align} & (ii)A-B=\left[ \begin{matrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \\ \end{matrix} \right]\Rightarrow \left( A-B \right)'=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right] \\ & A'-B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right] \\ \end{align}

Hence, we have verified that $\left( A-B \right)'=A'-B'$

3. If$\mathrm{A }\!\!'\!\!\text{ =}\left[ \begin{matrix} \mathrm{3} & \mathrm{4} \\ \mathrm{-1} & \mathrm{2} \\ \mathrm{0} & \mathrm{1} \\ \end{matrix} \right]$ and $\mathrm{B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \mathrm{1} & \mathrm{2} & \mathrm{3} \\ \end{matrix} \right]$ , then verify that \begin{align} & \mathrm{(i)}\left( \mathrm{A+B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ +B }\!\!'\!\!\text{ } \\ & \mathrm{(ii)}\left( \mathrm{A-B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ -B }\!\!'\!\!\text{ } \\ \end{align}

Ans:

We have

\begin{align} & A=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]B'=\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right] \\ & (i)A+B=\left[ \begin{matrix} 2 & 1 & 1 \\ 5 & 4 & 4 \\ \end{matrix} \right]\Rightarrow \left( A+B \right)'=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right] \\ & A'+B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right] \\ \end{align} Hence, we have verified that $\left( A+B \right)'=A'+B'$ \begin{align} & (ii)A-B=\left[ \begin{matrix} 4 & -3 & -1 \\ 3 & 0 & -2 \\ \end{matrix} \right]\Rightarrow \left( A-B \right)'=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right] \\ & A'-B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right] \\ \end{align} Hence, we have verified that $\left( A-B \right)'=A'-B'$

4. If$\mathrm{A }\!\!'\!\!\text{ =}\left[ \begin{matrix} \mathrm{-2} & \mathrm{3} \\ \mathrm{1} & \mathrm{2} \\ \end{matrix} \right]\mathrm{ }\!\!\And\!\!\text{ B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{0} \\ \mathrm{1} & \mathrm{2} \\ \end{matrix} \right]$ , then find $\left( \mathrm{A+2B} \right)\mathrm{ }\!\!'\!\!\text{ }$

Ans:

Solve for the condition\begin{align} & \therefore A=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right] \\ & \therefore A+2B=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+2\left[ \begin{matrix} -1 & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} -2 & 0 \\ 2 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} -4 & 1 \\ 5 & 6 \\ \end{matrix} \right] \\ & \therefore (A+2B)=\left[ \begin{matrix} -4 & 5 \\ 1 & 6 \\ \end{matrix} \right] \\ \end{align}

5. For the matrix $\mathrm{A }\;\;\And\;\;\text{ B}$ , verify that $\left( \mathrm{AB} \right)\mathrm{ }\!\!'\!\!\text{ =B }\!\!'\!\!\text{ A }\!\!'\!\!\text{ }$where

\begin{align} & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{1} \\ \mathrm{-4} \\ \mathrm{3} \\ \end{matrix} \right]\mathrm{,B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \end{matrix} \right] \\ & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{0} \\ \mathrm{1} \\ \mathrm{2} \\ \end{matrix} \right]\mathrm{,B=}\left[ \begin{matrix} \mathrm{1} & \mathrm{5} & \mathrm{7} \\ \end{matrix} \right] \\ \end{align}

Ans:

\begin{align} & (i)~AB=\left[ \begin{matrix} 1 \\ -4 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \\ \end{matrix} \right] \\ & \therefore (AB{)}'=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right] \\ & B'A'=\left[ \begin{matrix} -1 \\ 2 \\ 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -4 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right] \\ \end{align} Hence, we have verified $\left( AB \right)'=B'A'$ \begin{align} & (ii)~AB=\left[ \begin{matrix} 0 \\ 1 \\ 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \\ \end{matrix} \right] \\ & \therefore (AB{)}'=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right] \\ & B'A'=\left[ \begin{matrix} 1 \\ 5 \\ 7 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right] \\ \end{align}

Hence, we have verified $\left( AB \right)'=B'A'$

6. If \begin{align} & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{cos }\;\;\alpha\;\;\text{ } & \mathrm{sin }\;\;\alpha\!\!\text{ } \\ \mathrm{-sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\;\;\text{ } \\ \end{matrix} \right]\mathrm{, then verify thatA }\;\;\;\;\text{ A=I} \\ & \mathrm{(ii)A=}\left[ \begin{matrix} \mathrm{sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\!\!\text{ } \\ \mathrm{-cos }\;\;\alpha\!\!\text{ } & \mathrm{sin }\;\;\alpha\!\!\text{ } \\ \end{matrix} \right]\mathrm{, then verify thatA }\;\;'\;\;\text{ A=I} \\ \end{align}

Ans:

Solve for the condition

1. \begin{align} & A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & \therefore A'=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} {{\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{matrix} \right] \\ & A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I \\ \end{align} \begin{align} & A=\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & \therefore A'=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} {{\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{matrix} \right] \\ & A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I \\ \end{align}

7. (i) Show that the matrix $\mathrm{A=}\left[ \begin{matrix} \mathrm{1} & \mathrm{-1} & \mathrm{5} \\ \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \mathrm{5} & \mathrm{1} & \mathrm{3} \\ \end{matrix} \right]$ is a symmetric matrix (ii) Show that the matrix $\mathrm{A=}\left[ \begin{matrix} \mathrm{0} & \mathrm{1} & \mathrm{-1} \\ \mathrm{-1} & \mathrm{0} & \mathrm{1} \\ \mathrm{1} & \mathrm{-1} & \mathrm{0} \\ \end{matrix} \right]$ is a skew symmetric matrix

Ans:

(i) We have

\begin{align} & A'=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \\ \end{matrix} \right]=A \\ & \therefore A'=A \\ \end{align} Hence, A is a symmetric matrix (ii) We have \begin{align} & A'=\left[ \begin{matrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \\ \end{matrix} \right]=-\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]=-A \\ & \therefore A'=-A \\ \end{align}

Hence, A is a skew-symmetric matrix

8. For the matrix $\mathrm{A=}\left[ \begin{matrix} \mathrm{1} & \mathrm{5} \\ \mathrm{6} & \mathrm{7} \\ \end{matrix} \right]$, verify that (i) $\left( \mathrm{A+A }\!\!'\!\!\text{ } \right)$ is a symmetric matrix (ii) $\left( \mathrm{A+A }\!\!'\!\!\text{ } \right)$ is a skew symmetric matrix

Ans:

$A=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right],A'=\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]$ \begin{align} & (i)A+A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right] \\ & \left( A+A' \right)'=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right]=A+A' \\ \end{align} Hence, $\left( A+A' \right)$is a symmetric matrix \begin{align} & (ii)A-A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right] \\ & \left( A-A' \right)'=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} \right]=-\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]=A+A' \\ \end{align}

Hence, $\left( A-A' \right)$is a skew-symmetric matrix

9. Find $\frac{\mathrm{1}}{\mathrm{2}}\left( \mathrm{A+A }\;\;'\;\;\text{ } \right)\mathrm{ }\;\;\And\!\!\text{ }\frac{\mathrm{1}}{\mathrm{2}}\left( \mathrm{A-A }\;\;'\;\;\text{ } \right)$ , when $\mathrm{A=}\left[ \begin{matrix} \mathrm{0} & \mathrm{a} & \mathrm{b} \\ \mathrm{-a} & \mathrm{0} & \mathrm{c} \\ \mathrm{-b} & \mathrm{-c} & \mathrm{0} \\ \end{matrix} \right]$

Ans:

The given matrix is $A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]$, then $A'=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right]$ \begin{align} & \frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right] \\ \end{align} \begin{align} & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right] \\ \end{align}

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

$\left[ \begin{matrix} \mathrm{3} & \mathrm{5} \\ \mathrm{1} & \mathrm{-1} \\ \end{matrix} \right]$

Ans:

\begin{align} & (i)\frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 6 & 6 \\ 6 & -2 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\left[ \begin{matrix} 3 & 3 \\ 3 & -1 \\ \end{matrix} \right] \\ \end{align} Thus, $\frac{1}{2}\left( A+A' \right)$ is a symmetric matrix. \begin{align} & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & 4 \\ -4 & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right] \\ \end{align}

Thus, $\frac{1}{2}\left( A-A' \right)$ is a skew-symmetric matrix.

11. If A,B are symmetric matrix of same order, then $\mathrm{AB-BA}$ is a

1. Skew symmetric matrix

2. Symmetric matrix

3. Zero matrix

4. Identity matrix

Ans:

$A\And B$ are symmetric , therefore , we have

$A'=A\And B'=B$

Consider

\begin{align} & \left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)' \\ & =B'A'-A'B' \\ & =BA-AB \\ & =-\left( AB-BA \right) \\ & \therefore \left( AB-BA \right)'=-\left( AB-BA \right) \\ \end{align} Thus, $\left( AB-BA \right)'$ is a skew-symmetric matrix

12. If$\mathrm{A=}\left[ \begin{matrix} \mathrm{cos }\;\;\alpha\;\;\text{ } & \mathrm{-sin }\;\;\alpha\!\!\text{ } \\ \mathrm{sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\!\!\text{ } \\ \end{matrix} \right]$ , then $\mathrm{A+A }\;\;'\;\;\text{ =I}$ , if the value of $\mathrm{ }\;\;\alpha\!\!\text{ }$ is \begin{align} & \mathrm{A}\mathrm{.}\frac{\mathrm{ }\;\;\pi\;\;\text{ }}{\mathrm{6}} \\ & \mathrm{B}\mathrm{.}\frac{\mathrm{ }\;\;\pi\;\;\text{ }}{\mathrm{3}} \\ & \mathrm{C}\mathrm{. }\;\;\pi\!\!\text{ } \\ & \mathrm{D}\mathrm{.}\frac{\mathrm{3 }\;\;\pi\!\!\text{ }}{\mathrm{2}} \\ \end{align}

Ans:

\begin{align} & A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A+A'=I \\ & \therefore \left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]+\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & \alpha =\frac{\pi }{3} \\ \end{align}

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