The multiplicative inverse of matrix $\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$ is

  • A
    $\begin{bmatrix} 4 & -1 \\ -7 & -2 \end{bmatrix}$
  • B
    $\begin{bmatrix} -4 & -1 \\ 7 & -2 \end{bmatrix}$
  • C
    $\begin{bmatrix} 4 & -7 \\ 7 & 2 \end{bmatrix}$
  • D
    $\begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix}$

Explore More

Similar Questions

If $A$ and $B$ are square matrices of order $3$ such that $|A|=2$ and $|B|=4$,then $|A(\operatorname{adj} B)| = \dots$

Let $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = I + \operatorname{adj}(A) + (\operatorname{adj} A)^2 + \dots + (\operatorname{adj} A)^{10}$. Then,the sum of all the elements of the matrix $B$ is:

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]$

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ such that $A^2 - 4A + 3I = 0$,where $I$ is a unit matrix of order $2$,then $A^{-1}$ is

Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in R$. Then $(\alpha - \beta)^2$ 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