Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$,$\alpha \in R$ such that $A^{32} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Then a value of $\alpha$ is

  • A
    $0$
  • B
    $\frac{\pi}{16}$
  • C
    $\frac{\pi}{32}$
  • D
    $\frac{\pi}{64}$

Explore More

Similar Questions

Let $A = [a_{ij}]$ be a square matrix of order $2$ with entries either $0$ or $1$. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0$ for $(0 < x < \pi)$ are:

For a $3 \times 3$ matrix $A$,if $A(\operatorname{adj} A) = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 2 \\ 0 & 0 & -10 \end{bmatrix}$,then the value of the determinant of $A$ is:

Let $A = \begin{bmatrix} 5 & \sin^2 \theta & \cos^2 \theta \\ -\sin^2 \theta & -5 & 1 \\ \cos^2 \theta & 1 & 5 \end{bmatrix}$. Then the maximum value of $\det(A)$ is

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & -\sin t & \cos t \end{bmatrix}$. Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be the roots of $\det(A - \lambda I_{3}) = 0$,where $I_{3}$ denotes the identity matrix. If $\lambda_{1} + \lambda_{2} + \lambda_{3} = \sqrt{2} + 1$,then the set of possible values of $t$ for $-\pi \leq t < \pi$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo