If $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$, then the determinant of the matrix $(A^{2025} - 3A^{2024} + A^{2023})$ is

  • A
    $28$
  • B
    $12$
  • C
    $24$
  • D
    $16$

Explore More

Similar Questions

If $AA^T = I$ and $C$ is a skew-symmetric matrix,then $((A^T CA)^{50})^T$ is equal to

Let $\alpha \beta \gamma = 45$; $\alpha, \beta, \gamma \in R$. If $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in R$ such that $xyz \neq 0$,then $6\alpha + 4\beta + \gamma$ is equal to:

If $a, b, c$ are non-zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| \begin{array}{ccc} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ac & bc & a^2 + b^2 \end{array} \right| = k a^2 b^2 c^2$,then $k$ is equal to

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$, then among the two statements:
(S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A + B)) = -5$.

If one of the cube roots of $1$ be $\omega$, then $\left|\begin{array}{ccc}1 & 1+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|=$

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