If A = begin{bmatrix} i & 0 0 & i end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix}$.First,calculate $AB$:$AB = \begi

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$A^2 - B^2$

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(A) Given $A = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix}$.First,calculate $AB$:$AB = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix} = \begin{bmatrix} 0 & -i^2 \ -i^2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$.Next,calculate $BA$:$BA = \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix} \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} = \begin{bmatrix} 0 & -i^2 \ -i^2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$.Since $AB = BA$,the matrices $A$ and $B$ commute.For any two square matrices $A$ and $B$ that commute,the identity $(A + B)(A - B) = A^2 - AB + BA - B^2$ simplifies to $A^2 - B^2$ because $-AB + BA = 0$.Therefore,$(A + B)(A - B) = A^2 - B^2$.

If $A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}$,then $(A + B)(A - B)$ is equal to

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