Let $A, B, C$ be $3 \times 3$ non-singular matrices and $I$ be the identity matrix of order three. If $A B A = B A^2 B$ and $A^3 = I$, then $A B^4 - B^4 A = $

  • A
    $O_{3 \times 3}$
  • B
    $1/2$
  • C
    $1$
  • D
    $21$

Explore More

Similar Questions

Let $a_1, a_2, a_3, \dots, a_{10}$ be in $G.P.$ with $a_i > 0$ for $i = 1, 2, \dots, 10$ and $S$ be the set of pairs $(r, k)$,$r, k \in N$ (the set of natural numbers) for which
$\left| \begin{array}{ccc} \log_e(a_1^r a_2^k) & \log_e(a_2^r a_3^k) & \log_e(a_3^r a_4^k) \\ \log_e(a_4^r a_5^k) & \log_e(a_5^r a_6^k) & \log_e(a_6^r a_7^k) \\ \log_e(a_7^r a_8^k) & \log_e(a_8^r a_9^k) & \log_e(a_9^r a_{10}^k) \end{array} \right| = 0$
Then the number of elements in $S$ is:

Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if:

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where $bc \neq 0$) satisfies the equation $x^2 + k = 0$,then:

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 $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$,then show that $A^{3} - 23A - 40I = 0$.

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