If $A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$,then $|adj\,A|$ is equal to

  • A
    $16$
  • B
    $10$
  • C
    $6$
  • D
    None of these

Explore More

Similar Questions

If $A$ is a $2 \times 2$ order non-singular matrix,then the determinant of $A^{-1}$ is . . . . . . .

If $A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$,then $x =$ . . . . . . .

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 0 \\ -1 & 6 & 1 \end{bmatrix}$,then the adjoint of $A$ is

Let $X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,$Y=\alpha I+\beta X+\gamma X^{2}$ and $Z=\alpha^{2} I-\alpha \beta X+\left(\beta^{2}-\alpha \gamma\right) X^{2}$,where $\alpha, \beta, \gamma \in \mathbb{R}$. If $Y^{-1}=\begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$,then $(\alpha-\beta+\gamma)^{2}$ is equal to

If $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \cdot A \cdot \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then $A =$

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