A) \[3I\]
B) \[5I\]
C) \[-5I\]
D) None of these
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{A}^{2}}=\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 16+3 & 4+2 \\ 12+6 & 3+4 \\ \end{matrix} \right]=\left[ \begin{matrix} 19 & 6 \\ 18 & 7 \\ \end{matrix} \right]\] Now, \[{{A}^{2}}-6A=\left[ \begin{matrix} 19 & 6 \\ 18 & 7 \\ \end{matrix} \right]-6\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 19-24 & 6-6 \\ 18-18 & 7-12 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -5 & 0 \\ 0 & -5 \\ \end{matrix} \right]=-5\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=-5I\]
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