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

VSAT 2022

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.


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  • 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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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:

The correct answer is A

\[A\And B\] 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:

The correct answer is B

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


NCERT Solutions for Class 12 Maths Chapter 3 Other Exercises

Chapter 3 - Matrix Other Exercises in PDF Format

Exercise 3.1

10 Questions & Solutions (5 Short Answers, 5 Long Answers)

Exercise 3.2

22 Questions & Solutions (3 Short Answers, 19 Long Answers)

Exercise 3.4

18 Questions & Solutions (18 Short Answers)


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Students might refer to solutions for the last year or maybe some notes that are never updated. And during the exam time, you can’t manage time to find another set of solutions to prepare. Well, referring Vedantu NCERT Solutions keep you on the safer side as we make regular additions and modifications in the study material as per the changes in the NCERT guidelines.

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Referring to Vedantu’s easy and brief solutions will help you understand the problems in one go and retain it in your mind for a longer period. So, this is going to help you score good marks in the unit tests, half-yearly exams at school level and even board examinations. If you have difficulty in any of the chapters or are about to begin preparing for your board exams, school level exams, or JEE exam, you must start preparing for the right solutions today itself. To download chapter-wise or exercise-wise solutions, you can anytime download the free PDF file from Vedantu and begin preparing today. We even offer solutions to all other subjects as well.

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