If $A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]$,then find $A^8$. (in $B$)

  • A
    $4$
  • B
    $8$
  • C
    $64$
  • D
    $128$

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Similar Questions

If $I$ is the identity matrix of order $2$ and $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$,then for $n \geq 1$,mathematical induction gives:

Choose the correct option for the matrices given below:
$\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1\end{array}\right] \\ & B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right] \\ & C=\left[\begin{array}{ccc}\cos \frac{\pi}{6} & 0 & \sin \frac{\pi}{6} \\ 0 & 1 & 0 \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0\end{array}\right] \\ & D=\left[\begin{array}{ccc}\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\ -\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{array}\right]\end{aligned}$

If $A=\left[\begin{array}{rrr}0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0\end{array}\right]$,$B=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right]$,and $C=\left[\begin{array}{c}2 \\ -2 \\ 3\end{array}\right]$. Calculate $AC$,$BC$,and $(A + B)C$. Also,verify that $(A+B)C = AC + BC$.

If $A$ is an $m \times n$ matrix and $B$ is a matrix such that both $AB$ and $BA$ are defined,then the order of $B$ is

$AB = 0$,if and only if

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