NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3

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:

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

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

Conclusion

In conclusion, class 12 ex 3.3 has thoroughly explored the fundamental concepts of matrices, including various operations, types, properties, and their applications. This exercise is essential for understanding how matrices are used in solving linear equations and performing geometric transformations. Mastery of these concepts is crucial for students aiming to excel in advanced mathematics and related fields.

Class 12 Maths Chapter 3: Exercises Breakdown

Other Study Materials for CBSE Class 12 Maths Chapter 3 Matrices

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths