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#### Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 11

Answer: $A^{2}=\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right]$

Hint:

matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

Consider,

$A^{2}=A A\\\\ A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right] \\\\ \\ =\left[\begin{array}{cc}(\cos 2 \theta) \times(\cos 2 \theta)+(\sin 2 \theta) \times(-\sin 2 \theta) & \cos 2 \theta \times \sin 2 \theta+\sin 2 \theta \times \cos 2 \theta \\ (-\sin 2 \theta) \times(\cos 2 \theta)+\cos 2 \theta \times(-\sin 2 \theta) & \sin 2 \theta \times(-\sin 2 \theta)+\cos 2 \theta \times \cos 2 \theta\end{array}\right]\\\\\\ =\left[\begin{array}{cc}\cos ^{2} 2 \theta-\sin ^{2} 2 \theta & \cos 2 \theta \sin 2 \theta+\sin 2 \theta \cos 2 \theta \\ -\cos 2 \theta \sin 2 \theta-\sin 2 \theta \cos 2 \theta & -\sin ^{2} 2 \theta+\cos ^{2} 2 \theta\end{array}\right]$

We know that,  $\cos ^{2} \theta-\sin ^{2} \theta=\cos ^{2} 2 \theta$

$A^{2}=\left[\begin{array}{cc} \cos (2 \times 2 \theta) & 2 \sin 2 \theta \cos 2 \theta \\ -2 \sin 2 \theta \cos 2 \theta & \cos (2 \times 2 \theta) \end{array}\right]$

Again, we have

$\begin{array}{l} \\\\ \sin 2 \theta=2 \sin \theta \cos \theta\\\\ A^{2}=\left[\begin{array}{cc} \cos 4 \theta & \sin (2 \times 2 \theta) \\ -\sin (2 \times 2 \theta) & \cos 4 \theta \end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc} \cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta \end{array}\right] \end{array}$