If A = begin{bmatrix} 2i & i^3 i^2 & 1 end{b | vedclass

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(A) Given $A = \begin{bmatrix} 2i & i^3 \ i^2 & 1 \end{bmatrix}$. Since $i^2 = -1$ and $i^3 = -i$, we have $A = \begin{bmatrix} 2i & -i \ -1 & 1 \en

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$\begin{bmatrix} -i & 1 \ -i & 2 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} 2i & i^3 \ i^2 & 1 \end{bmatrix}$. Since $i^2 = -1$ and $i^3 = -i$, we have $A = \begin{bmatrix} 2i & -i \ -1 & 1 \end{bmatrix}$.Step $1$: Find the determinant $|A| = (2i)(1) - (-i)(-1) = 2i - i = i$.Step $2$: Find the adjoint $adj(A)$. For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$, $adj(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus, $adj(A) = \begin{bmatrix} 1 & i \ 1 & 2i \end{bmatrix}$.Step $3$: Calculate $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{i} \begin{bmatrix} 1 & i \ 1 & 2i \end{bmatrix}$.Since $\frac{1}{i} = -i$, we get $A^{-1} = -i \begin{bmatrix} 1 & i \ 1 & 2i \end{bmatrix} = \begin{bmatrix} -i & -i^2 \ -i & -2i^2 \end{bmatrix} = \begin{bmatrix} -i & 1 \ -i & 2 \end{bmatrix}$.

If $A = \begin{bmatrix} 2i & i^3 \\ i^2 & 1 \end{bmatrix}$, then $A^{-1}$ is equal to

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