A) \[\left[ \begin{matrix} 1 \\ 0 \\ 2 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \frac{8}{3} \\ \frac{-1}{3} \\ 0 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \frac{-8}{3} \\ 1 \\ 0 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} \frac{8}{3} \\ \frac{1}{3} \\ -1 \\ \end{matrix} \right]\]
E) \[\left[ \begin{matrix} \frac{8}{3} \\ \frac{1}{3} \\ 0 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
\[\because \] \[A=\left| \begin{matrix} 1 & -1 & -2 \\ 2 & 1 & 1 \\ 4 & -1 & -2 \\ \end{matrix} \right|\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{3}\left[ \begin{matrix} -1 & 0 & 1 \\ 8 & 6 & -5 \\ -6 & -3 & 3 \\ \end{matrix} \right]\] Now, \[{{A}^{-1}}D=\frac{1}{3}\left[ \begin{matrix} -1 & 0 & 1 \\ 8 & 6 & -5 \\ -6 & -3 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 3 \\ 5 \\ 11 \\ \end{matrix} \right]\] \[=\frac{1}{3}\left[ \begin{matrix} 8 \\ -1 \\ 0 \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8/3 \\ -1/3 \\ 0 \\ \end{matrix} \right]\]
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