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  4. Inverse of the matrix [ cos2θ&- sin2θ sin2θ cos2θ ] is :

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Q. Inverse of the matrix $\begin{bmatrix}\cos2\theta&-\sin2\theta \\ \sin2\theta &\cos2\theta \end{bmatrix}$ is :

Determinants

Solution:

Let, $A = \begin{bmatrix}\cos2\theta&-\sin2\theta \\ \sin2\theta &\cos2\theta \end{bmatrix} $
And $A^{-1} = \frac{Adj A}{\left|A\right|} $
Here $\left|A\right| = \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{vmatrix}A_{11}&A_{12}\\ A_{21}&A_{22}\end{vmatrix} $
Where, $A_{11}$ = cofactor
and , $A_{11} = (-1)^{1+} . \cos 2 \theta = \cos 2 \theta $
$A_{12} = (-1)^{1+2} . \sin 2 \theta = - \sin 2 \theta $
$A_{21} = (-1)^{2+1} . (- \sin 2\theta) = + \sin 2 \theta$
$A_{22} = (- 1)^{2+2} \cos 2 \theta = \cos 2 \theta$
Hence, Adj $A = \begin{bmatrix}\cos2\theta&-\sin2\theta \\ \sin2\theta &\cos2\theta \end{bmatrix}^T$
Where 'T' denotes the transpose of the matrix. And the transpose of the matrix is obtained by interchanging the rows and columns of the given matrix.
Thus, Adj $(A) = \begin{bmatrix}\cos2\theta&\sin2\theta \\ - \sin2\theta &\cos2\theta \end{bmatrix}$
$\Rightarrow \ A^{-1} = \begin{bmatrix}\cos2\theta& \sin2\theta \\ - \sin2\theta &\cos2\theta \end{bmatrix}$