A) \[{A}'=\left[ \begin{matrix} 1 & \,\,1 \\ 1 & -1 \\ \end{matrix} \right]\]
B) \[{{A}^{-1}}=\left[ \begin{matrix} \,\,1 & 1 \\ -1 & 1 \\ \end{matrix} \right]\]
C) \[A.\,\,\left[ \begin{matrix} \,\,1 & 1 \\ -1 & 1 \\ \end{matrix} \right]=2I\]
D) \[\lambda A=\left[ \begin{matrix} \lambda & -\lambda \\ 1 & -1 \\ \end{matrix} \right]\]where \[\lambda \]is a non zero scalar
Correct Answer: C
Solution :
\[A\,.\,\left[ \begin{matrix} 1 & 1 \\ -1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 1 \\ -1 & 1 \\ \end{matrix} \right]\]=\[\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right]=2\,\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=2I\].
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