A) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\]
D) \[-\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
[a] \[\left[ \begin{matrix} 2 & 1 \\ 3 & 2 \\ \end{matrix} \right]A\left[ \begin{matrix} -3 & 2 \\ 5 & -3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[A={{\left[ \begin{matrix} 2 & 1 \\ 3 & 2 \\ \end{matrix} \right]}^{-1}}\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]{{\left[ \begin{matrix} -3 & 1 \\ 5 & -3 \\ \end{matrix} \right]}^{-1}}\] \[=\left[ \begin{matrix} 2 & -1 \\ -3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -3 & -2 \\ -5 & -3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2 & -1 \\ -3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\ 5 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
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