A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 nor 2
Correct Answer: B
Solution :
[b] \[A=\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\] \[A.A=\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2 & -2 \\ -2 & 2 \\ \end{matrix} \right]=-2\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\] \[{{A}^{2}}=-2A\] \[{{A}^{2}}.A=-2\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\] \[=-2\left[ \begin{matrix} 2 & -2 \\ -2 & 2 \\ \end{matrix} \right]=4\left[ \begin{matrix} -1 & 1 \\ 1 & -1 \\ \end{matrix} \right]\] \[{{A}^{3}}=4A\] Hence \[{{A}^{2}}\ne -A,{{A}^{3}}=4A\]
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