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  4. Inverse of the matrix [ cos 2 θ - sin 2 θ sin 2 θ cos 2 θ] is

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Q. Inverse of the matrix $\begin{bmatrix}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{bmatrix}$ is

Determinants

Solution:

Let, $ A =\begin{bmatrix} cos\, 2 \theta & -sin \,2 \theta \\
sin\, 2 \theta & cos \,2 \theta \end{bmatrix} $
And $A ^{-1}=\frac{Adj. A}{|A|}$
Here $| A |=cos ^{2} 2 \theta-\left(-\sin ^{2} 2 \theta\right)=\cos ^{2} 2 \theta+\sin ^{2} 2 \theta$
$=1 \left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$
And, Adj $A =\begin{bmatrix} A _{11} & A _{12} \\ A _{21} & A _{22}
\end{bmatrix} \\$
Where, $ A _{11}$ cofactor
and, $A_{11}=(-1)^{i+i} \cdot \cos 2 \theta=\cos 2 \theta$
$A_{12}=(-1)^{1+2} \cdot \sin 2 \theta=-\sin 2 \theta$
$A_{21}=(-1)^{2+1} \cdot(-\sin 2 \theta)=+\sin 2 \theta$
$A_{22}=(-1)^{2+2} \cos 20=\cos 20$
Hence, Adj A $=\begin{bmatrix}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{bmatrix}^{ T }$
Thus, Adj ( A ) $=\begin{bmatrix}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{bmatrix}$
$\Rightarrow A ^{-1}=\begin{bmatrix}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{bmatrix}$