A) \[-14I\]
B) \[-10I\]
C) \[8I\]
D) \[-\frac{1}{14}I\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 3 & 2 \\ 1 & -4 \\ \end{matrix} \right]\] \[adj(A)=\left[ \begin{matrix} -4 & -2 \\ -1 & 3 \\ \end{matrix} \right]\] \[\therefore \] \[A(adjA)=\left[ \begin{matrix} 3 & 2 \\ 1 & -4 \\ \end{matrix} \right]\left[ \begin{matrix} -4 & -2 \\ -1 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -12-2 & -6+6 \\ -4+4 & -2-12 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -14 & 0 \\ 0 & -14 \\ \end{matrix} \right]\] \[=-14\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=-14I\]
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