Let $\mathrm{A}=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right),(\alpha \in \mathrm{R})$ such that $\mathrm{A}^{32}=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)$. Then a value of $\alpha$ is:

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Matrices

Let

A = (cos α sin α - sin α cos α), (α \in R)

such that

A^{32} = (01 - 1 0)

. Then a value of

α

is:

Answer: (D)

The matrix is a rotation by

α

, and powers of a rotation simply add the angles, so

A^{32}

is a rotation by

32 α

, namely

(cos 32 α sin 32 α - sin 32 α cos 32 α)

. Matching this to

(01 - 1 0)

needs

cos 32 α = 0

and

sin 32 α = 1

, so

32 α = \frac{π}{2} + 2 nπ

. Taking

n = 0

gives

α = \frac{π}{64}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option