First slide

Operations of matrix

Question

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

Moderate

A

$128 B$
B

$32 B$
C

$16 B$
D

$64 B$

Solution

$A = i B$
$\begin{array}{l} \Rightarrow A^{2} = i^{2} B^{2} = (- 1) (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array}) (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array}) = - 2 B \\ \Rightarrow A^{4} = (- 2 B)^{2} = 4 B^{2} = 4 (2 B) = 8 B \\ \Rightarrow A^{8} = (8 B)^{2} = 64 B^{2} = 64 (2 B) = 128 B \end{array}$

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