Let $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$. If $(A B^{-1})^{-1} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $2b + 5c + 10d =$

  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • D
    $2$

Explore More

Similar Questions

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,show that $A^{2} - 5A + 7I = 0$. Hence,find $A^{-1}$.

If $A=\left[\begin{array}{ll}2 & -2 \\ 2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$,then $(B^{-1} A^{-1})^{-1} = ?$

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$,then $\alpha^2 + k^2$ is equal to?

Matrix $A = \begin{bmatrix} 1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1 \end{bmatrix}$ is invertible for

If $A^T$ denotes the transpose of the matrix $A = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix}$,where $a, b, c, d, e$ and $f$ are integers such that $abd \neq 0$,then the number of such matrices for which $A^{-1} = A^T$ 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