If $A = \begin{bmatrix} i & - i \ - i & i \end{bmatrix}$ and $B = \ \begin{bmatrix} 1 & - 1 \ - 1 & 1 \end{bmatrix}$, then $A^{8}$ equals

Question

Dr. Abubacker · Accepted Answer

(b)
We have,
$A = iB$
$\Rightarrow A^{2} = (iB)^{2} = i^{2}B^{2} = - B^{2} = \begin{bmatrix}
2 & - 2 \
 - 2 & 2
\end{bmatrix} = - 2B$
$\Rightarrow A^{4} = ( - 2B)^{2} = 4B^{2} = 4(2B) = 8B$
$\Rightarrow \left( A^{4} ight)^{2} = (8B)^{2}$
$\Rightarrow A^{8} = 64B^{2} = 128B$

Q. If $A = \begin{bmatrix} i & - i \\ - i & i \end{bmatrix}$ and $B = \ \begin{bmatrix} 1 & - 1 \\ - 1 & 1 \end{bmatrix}$, then $A^{8}$ equals

Explanation

Reviewed by Dr. Abubacker

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