The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$ is

  • A
    $-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 2 & -3 \end{bmatrix}$
  • B
    $-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$
  • C
    $-\frac{1}{3} \begin{bmatrix} 3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & -3 \end{bmatrix}$
  • D
    $-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

Explore More

Similar Questions

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

If possible,using elementary row transformations,find the inverse of the following matrix:
$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

If $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$ where $i = \sqrt{-1}$ and $B = A^{2029}$,then $B^{-1} =$

If $A$ is a matrix of order $3$,such that $A(\operatorname{adj} A) = 10I$,then $|\operatorname{adj} A| = $

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