A) \[\left[ \begin{matrix} -3 & 4 \\ -4 & 13 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 3 & 4 \\ 14 & 13 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -3 & 4 \\ 14 & -13 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 3 & -4 \\ -14 & 13 \\ \end{matrix} \right]\]
Correct Answer: C
Solution :
We have \[\left[ \begin{matrix} 3 & 1 \\ 4 & 1 \\ \end{matrix} \right]X=\left[ \begin{matrix} 5 & -1 \\ 2 & 3 \\ \end{matrix} \right]\] ?(i) Let \[A=\left[ \begin{matrix} 3 & 1 \\ 4 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A|=3-4=-1\] \[\text{adj}\,\text{A}=\,\left[ \begin{matrix} 1 & -1 \\ -4 & 3 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{adjA}{|A|}=\frac{\left[ \begin{matrix} 1 & -1 \\ -4 & 3 \\ \end{matrix} \right]}{-1}\] \[=\left[ \begin{matrix} -1 & 1 \\ 4 & -3 \\ \end{matrix} \right]\] Now, \[AX=\left[ \begin{matrix} 5 & -1 \\ 2 & 3 \\ \end{matrix} \right]\] [from (i)] \[\Rightarrow \] \[X={{A}^{-1}}\left[ \begin{matrix} 5 & -1 \\ 2 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -1 & 1 \\ 4 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & -1 \\ 2 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -3 & 4 \\ 14 & -13 \\ \end{matrix} \right]\]You need to login to perform this action.
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