If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A^{-1} = A'$, then find the value of $\alpha$.

$\alpha \in \mathbb{R}$ (Any real number)

The correct answer is Option (2) → $\alpha \in \mathbb{R}$ (Any real number) ##

We have,

$A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \text{ and } A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Also, $A^{-1} = A'$

$\Rightarrow AA^{-1} = AA'$

$\Rightarrow I = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

$I = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha \\ -\sin \alpha \cdot \cos \alpha + \cos \alpha \cdot \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix}$

Since, both matrices are equal.

$∴$ On comparing the corresponding elements of both sides, we get

$\cos^2 \alpha + \sin^2 \alpha = 1$

which is true for all real values of $\alpha$.