• # question_answer Let$X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right],A=\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \\ \end{matrix} \right]$and$B=\left[ \begin{matrix} 3 \\ 1 \\ 4 \\ \end{matrix} \right]$. If$AX=B,$then$X$is equal to: 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)  $\left[ \begin{matrix} 0 \\ 2 \\ 1 \\ \end{matrix} \right]$

$\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,     ${{A}^{-1}}B=\frac{1}{5}\left[ \begin{matrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 \\ 1 \\ 4 \\ \end{matrix} \right]$                 $\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} -1 \\ 2 \\ 3 \\ \end{matrix} \right]$