A) \[{{A}^{2}}=A\]
B) \[{{B}^{2}}=B\]
C) \[AB\ne BA\]
D) \[AB=BA\]
Correct Answer: C
Solution :
Since \[{{A}^{2}}=\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} -4 & 2 \\ -3 & -6 \\ \end{matrix} \right]\ne A\] \[{{B}^{2}}=\left[ \begin{matrix} -1 & 0 \\ 2 & 3 \\ \end{matrix} \right]\,\left[ \begin{matrix} -1 & 0 \\ 2 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 4 & 9 \\ \end{matrix} \right]\ne B\] Now \[AB=\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ \end{matrix} \right]\,\left[ \begin{matrix} -1 & 0 \\ 2 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 6 \\ 3 & 0 \\ \end{matrix} \right]\] and \[BA=\left[ \begin{matrix} -1 & 0 \\ 2 & 3 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -2 \\ -7 & 4 \\ \end{matrix} \right]\] Obviously, \[AB\ne BA\].
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