If the matrix $A =\begin{bmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{bmatrix}$, then $A^2$ is equal to:
Answer & explanation
Correct answer: option 1
$A =\begin{bmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{bmatrix}$
so, $A^2=A×A$
$=\begin{bmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{bmatrix}\begin{bmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{bmatrix}=\begin{bmatrix}\cos^2 θ-\sin^2 θ&2\cos θ\sin θ\\-2\cos θ\sin θ&\cos^2 θ-\sin^2 θ\end{bmatrix}$
$=\begin{bmatrix}\cos 2θ&\sin 2θ\\-\sin 2θ&\cos 2θ\end{bmatrix}$ $[∵\cos^2 θ-\sin^2 θ=\cos 2θ, 2\cos θ\sin θ=\sin 2θ]$