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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 $C=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$, then
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Verified Answer
The correct answer is:
$\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=3 \mathrm{~A}^2 \mathrm{~B}^2 \mathrm{C}^2$
$$
\begin{aligned}
& \text { } \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right], \mathrm{B}^2=\mathrm{B} \cdot \mathrm{B}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \\
& \mathrm{C}^2=\mathrm{C} \cdot \mathrm{C}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]
\end{aligned}
$$
Hence,
$$
\begin{aligned}
& A^2+B^2+C^2=\left[\begin{array}{cc}
-3 & 0 \\
0 & -3
\end{array}\right] \\
& \text { and } 3 \mathrm{~A}^2 \mathrm{~B}^2 \mathrm{C}^2=3\left[\left(\mathrm{~A}^2 \mathrm{~B}^2\right) \mathrm{C}^2\right] \\
& =3\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]=\left[\begin{array}{cc}
-3 & 0 \\
0 & -3
\end{array}\right] \\
& \therefore \mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=3 \mathrm{~A}^2 \mathrm{~B}^2 \mathrm{C}^2 \\
&
\end{aligned}
$$
\begin{aligned}
& \text { } \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right], \mathrm{B}^2=\mathrm{B} \cdot \mathrm{B}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \\
& \mathrm{C}^2=\mathrm{C} \cdot \mathrm{C}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]
\end{aligned}
$$
Hence,
$$
\begin{aligned}
& A^2+B^2+C^2=\left[\begin{array}{cc}
-3 & 0 \\
0 & -3
\end{array}\right] \\
& \text { and } 3 \mathrm{~A}^2 \mathrm{~B}^2 \mathrm{C}^2=3\left[\left(\mathrm{~A}^2 \mathrm{~B}^2\right) \mathrm{C}^2\right] \\
& =3\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]=\left[\begin{array}{cc}
-3 & 0 \\
0 & -3
\end{array}\right] \\
& \therefore \mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=3 \mathrm{~A}^2 \mathrm{~B}^2 \mathrm{C}^2 \\
&
\end{aligned}
$$
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