If A=left[begin{array}{cc}i & 0 0 & -iend{ar | vedclass

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(A) Given matrices are $A=\left[\begin{array}{cc}i & 0 \ 0 & -i\end{array}ight]$,$B=\left[\begin{array}{cc}0 & -1 \ 1 & 0\end{array}ight]$,and $

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$A^2+B^2+C^2=3 A^2 B^2 C^2$

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(A) Given matrices are $A=\left[\begin{array}{cc}i & 0 \ 0 & -i\end{array}ight]$,$B=\left[\begin{array}{cc}0 & -1 \ 1 & 0\end{array}ight]$,and $C=\left[\begin{array}{cc}0 & i \ i & 0\end{array}ight]$.Calculating the squares of the matrices:$A^2 = \left[\begin{array}{cc}i & 0 \ 0 & -i\end{array}ight] \left[\begin{array}{cc}i & 0 \ 0 & -i\end{array}ight] = \left[\begin{array}{cc}i^2 & 0 \ 0 & (-i)^2\end{array}ight] = \left[\begin{array}{cc}-1 & 0 \ 0 & -1\end{array}ight] = -I$.$B^2 = \left[\begin{array}{cc}0 & -1 \ 1 & 0\end{array}ight] \left[\begin{array}{cc}0 & -1 \ 1 & 0\end{array}ight] = \left[\begin{array}{cc}-1 & 0 \ 0 & -1\end{array}ight] = -I$.$C^2 = \left[\begin{array}{cc}0 & i \ i & 0\end{array}ight] \left[\begin{array}{cc}0 & i \ i & 0\end{array}ight] = \left[\begin{array}{cc}i^2 & 0 \ 0 & i^2\end{array}ight] = \left[\begin{array}{cc}-1 & 0 \ 0 & -1\end{array}ight] = -I$.Summing the squares:$A^2+B^2+C^2 = (-I) + (-I) + (-I) = -3I = \left[\begin{array}{cc}-3 & 0 \ 0 & -3\end{array}ight]$.Now,calculating $3 A^2 B^2 C^2$:$3 A^2 B^2 C^2 = 3(-I)(-I)(-I) = 3(-I)^3 = 3(-I) = -3I$.Since $A^2+B^2+C^2 = -3I$ and $3 A^2 B^2 C^2 = -3I$,we conclude that $A^2+B^2+C^2 = 3 A^2 B^2 C^2$.

If $A=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$,$B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$ and $C=\left[\begin{array}{cc}0 & i \\ i & 0\end{array}\right]$,then which of the following is true?

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