If $A = \begin{bmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{bmatrix}$ and $\det\left(A^{2} - \frac{1}{2} I\right) = 0$,then a possible value of $\alpha$ is

  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Explore More

Similar Questions

Let $A$ be a $3 \times 3$ invertible matrix. If $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$,then $|A^2|$ is equal to

If $a, b, c$ and $d$ are complex numbers,then the determinant $\Delta = \begin{vmatrix} 2 & a+b+c+d & ab+cd \\ a+b+c+d & 2(a+b)(c+d) & ab(c+d)+cd(a+b) \\ ab+cd & ab(c+d)+cd(a+b) & 2abcd \end{vmatrix}$ is

Difficult
View Solution

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$ satisfies the equation $A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ for some real numbers $\alpha$ and $\beta$,then $\beta - \alpha$ is equal to ........ .

Let $|M|$ denote the determinant of a square matrix $M$. Let $g:\left[0, \frac{\pi}{2}\right] \rightarrow R$ be the function defined by $g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$,where $f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$. Let $p(x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$,and $p(2)=2-\sqrt{2}$. Then,which of the following is/are $TRUE$?
$(A) \ p \left(\frac{3+\sqrt{2}}{4}\right) < 0$
$(B) \ p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
$(C) \ p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
$(D) \ p \left(\frac{5-\sqrt{2}}{4}\right) < 0$

Matrix $A$ is such that ${A^2} = 2A - I$,where $I$ is the identity matrix. Then for $n \ge 2$,${A^n} = $

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