A) \[\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 0 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Given A\[\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{matrix} \right]\] Applying \[{{C}_{1}}\leftrightarrow {{C}_{3}}\] \[A\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] pre-multiplying both sides by \[{{A}^{-1}}\] \[{{A}^{-1}}A\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \\ \end{matrix} \right]={{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[I\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \\ \end{matrix} \right]={{A}^{-1}}I={{A}^{-1}}\] (\[\because \]\[{{A}^{-1}}A\]= I and I = Identity matrix) Hence,\[{{A}^{-1}}=\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 2 \\ \end{matrix} \right]\]
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