If A = begin{bmatrix} i & 0 0 & i/2 end{bmat | vedclass

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(B) Given $A = \begin{bmatrix} i & 0 \ 0 & i/2 \end{bmatrix}$.The determinant of $A$ is $|A| = (i)(i/2) - (0)(0) = i^2/2 = -1/2$.The adjoint of a $2

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$\begin{bmatrix} -i & 0 \ 0 & -2i \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} i & 0 \ 0 & i/2 \end{bmatrix}$.The determinant of $A$ is $|A| = (i)(i/2) - (0)(0) = i^2/2 = -1/2$.The adjoint of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$adj(A) = \begin{bmatrix} i/2 & 0 \ 0 & i \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A)$.$A^{-1} = \frac{1}{-1/2} \begin{bmatrix} i/2 & 0 \ 0 & i \end{bmatrix} = -2 \begin{bmatrix} i/2 & 0 \ 0 & i \end{bmatrix} = \begin{bmatrix} -i & 0 \ 0 & -2i \end{bmatrix}$.

If $A = \begin{bmatrix} i & 0 \\ 0 & i/2 \end{bmatrix}$ where $i = \sqrt{-1}$,then $A^{-1} = $

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