If A - - i -i- -i i - and B - - 1 -1- -1 1 - then A8 equals

We have A = iB ⇒ A^2 = (iB)^2 = i^2B^2 = B^2 = [ 2 amp; 2; 2 amp; 2 ]= 2B ⇒ A^4 = ( 2B)^2 = 4B^2 = 4(2B) = 8B ⇒ (A^4)^2 = (8B)^2 ⇒ A^8 = 64B ...

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Multiplication of Matrices

If A = [ i -i...

Question

$I f A = [\begin{matrix} i & - i - i & i \end{matrix}] a n d B = [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}], t h e n A^{8} e q u a l s$

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128 B

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- 128 B

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- 64 B

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A = [\begin{matrix} i & - i - i & i \end{matrix}], B [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}] \Rightarrow A^{8} =

A = (\begin{matrix} i & - i - i & i \end{matrix})

B = (\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}),

A^{8}

A = [\begin{matrix} i & - i - i & i \end{matrix}], i = \sqrt{- 1}

A^{8} [\begin{matrix} x y \end{matrix}] = [\begin{matrix} 864 \end{matrix}]

A = \frac{1}{\sqrt{3}} [\begin{matrix} 1 & i + 1 i - 1 & 1 \end{matrix}]

A (^{T})

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