Compute the indicated products $\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
- A$\left[\begin{array}{ccc}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]$
- B$\left[\begin{array}{ccc}3 & 4 & -1 \\ -8 & -13 & -9\end{array}\right]$
- C$\left[\begin{array}{ccc}-3 & 4 & 1 \\ 8 & -13 & 9\end{array}\right]$
- D$\left[\begin{array}{ccc}3 & -4 & 1 \\ -8 & 13 & -9\end{array}\right]$
Explore More
Similar Questions
If $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} = 0$,then $2x + 9 =$ . . . . . .
Let $A = [a_{ij}]$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$,$A \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$,and $A \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$. Then $a_{23}$ equals:
DifficultJEE MAIN 2025
View SolutionLet $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$. The only correct statement about the matrix $A$ is:
$\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & -1 \end{bmatrix} = $
Easy
View SolutionIf $A = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix}$ 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:
DifficultAIEEE 2012
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo