If $A$ is a square matrix of order $3$ with $|A| = 2$,then the value of $|(A - A^T)^5| + |(A^T - A)^3|$ is-

  • A
    $24$
  • B
    $16$
  • C
    $0$
  • D
    $8$

Explore More

Similar Questions

If $P$ and $Q$ are two non-singular matrices of the same order such that $Q^r = I$,for some integer $r > 1$,then $P^{-1}Q^{r-1}P - P^{-1}Q^{-1}P$ is equal to (where $I$ is the identity matrix and $O$ is the null matrix).

Let $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}$. For $k \in N$,if $X^{T} A^{k} X = 33$,then $k$ is equal to:

Let $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$,where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $xyz = 2$. If $A^2 = I_3$,then the value of $x^3 + y^3 + z^3$ is ............

Let integers $a, b \in [-3, 3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs $(a, b)$,for which $|\frac{z-a}{z+b}|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$ for some $z \in \mathbb{C}$,where $\omega$ and $\omega^2$ are the roots of $x^2+x+1=0$,is equal to . . . . . .

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B = I$,where $I$ is the identity matrix and $B$ is any square matrix of the same order as $A$. The matrix product $AB$ is equal to:

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