A) \[\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} -1 \\ -2 \\ 3 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -1 \\ -2 \\ -3 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} -1 \\ 2 \\ 3 \\ \end{matrix} \right]\]
E) \[\left[ \begin{matrix} 0 \\ 2 \\ 1 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
\[\because \]\[A=\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{5}\left[ \begin{matrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \\ \end{matrix} \right]\] Now, \[{{A}^{-1}}B=\frac{1}{5}\left[ \begin{matrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 \\ 1 \\ 4 \\ \end{matrix} \right]\] \[\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} -1 \\ 2 \\ 3 \\ \end{matrix} \right]\]You need to login to perform this action.
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