A) 0
B) \[\pm \text{ }1\]
C) - 1
D) 1
Correct Answer: B
Solution :
\[{{A}^{2}}=A\,.\,A=\left[ \begin{matrix} \lambda & 1 \\ -1 & -\lambda \\ \end{matrix} \right]\,\left[ \begin{matrix} \lambda & 1 \\ -1 & -\lambda \\ \end{matrix} \right]=\left[ \begin{matrix} {{\lambda }^{2}}-1 & 0 \\ 0 & -1+{{\lambda }^{2}} \\ \end{matrix} \right]=0\](As given) Þ \[{{\lambda }^{2}}-1=0\Rightarrow \lambda =\pm 1\].
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