If $A = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix}$ and $I$ is a unit matrix of $3^{rd}$ order,then $(A^2 + 9I)$ equals

  • A
    $2A$
  • B
    $4A$
  • C
    $6A$
  • D
    None of these

Explore More

Similar Questions

If $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ 1 & -2 \\ 0 & 3 \end{bmatrix}$,then $AB =$ . . . . . . .

Find the values of $x$ and $y$ from the following equation:
$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

Compute the indicated product: $\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right] \times \left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$

If $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(kB)^{\prime} = kB^{\prime}$,where $k$ is any constant.

If $A = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \\ 0 \\ 0 \end{bmatrix}$ are two matrices,then $AB^T$ is a non-zero matrix for $|\alpha|$ not 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