A) \[\left[ \begin{matrix} -5 & 1 \\ -2 & 2 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} -5 & -2 \\ 1 & 2 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -5 & -2 \\ 2 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 5 & 2 \\ -2 & -1 \\ \end{matrix} \right]\]
Correct Answer: C
Solution :
[c] Let \[\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]=B\] Then \[BA=\left[ \begin{matrix} -1 & 0 \\ 6 & 3 \\ \end{matrix} \right]\] \[\Rightarrow A={{B}^{-1}}\left[ \begin{matrix} -1 & 0 \\ -6 & 3 \\ \end{matrix} \right]\] \[\left| B \right|=3,\] adj \[B=\left[ \begin{matrix} 3 & -2 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{B}^{-1}}=\frac{1}{3}\left[ \begin{matrix} 3 & -2 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow A=\] \[\frac{1}{3}\left[ \begin{matrix} 3 & -2 \\ 0 & 3 \\ \end{matrix} \right]=\frac{1}{3}\left[ \begin{matrix} -3-12 & -6 \\ 6 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -5 & -2 \\ 2 & 1 \\ \end{matrix} \right]\]
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