If $A=\begin{bmatrix}\cos θ &\sin θ\\-\sin θ&\cos θ\end{bmatrix}$, then $A^2 = 1$, is true for
Answer & explanation
Correct answer: option 1
We have,
$A^2=\begin{bmatrix}\cos 2θ &\sin 2θ\\-\sin 2θ&\cos 2θ\end{bmatrix}$
$∴A^2=I= \cos 2θ=1$ and $\sin 2θ=0⇒ θ=0$.