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

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

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