A) 2A
B) 2B
C) \[2I\]
D) 0
Correct Answer: D
Solution :
\[A=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix} 0 & -i \\ -i & 0 \\ \end{matrix} \right]\] \[\therefore \] \[A+B=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -i \\ -i & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} i & -i \\ -i & i \\ \end{matrix} \right]\] \[A-B=\left[ \begin{matrix} i & 0 \\ 0 & i \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & i \\ -i & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} i & i \\ i & i \\ \end{matrix} \right]\] Now, \[(A+B)(A-B)\] \[=\left[ \begin{matrix} i & -i \\ -i & i \\ \end{matrix} \right]\left[ \begin{matrix} i & i \\ i & i \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{i}^{2}}-{{i}^{2}} & {{i}^{2}}-{{i}^{2}} \\ -{{i}^{2}}+{{i}^{2}} & -{{i}^{2}}+{{i}^{2}} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -1+1 & -1+1 \\ 1-1 & 1-1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]=0\]
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