A) \[\left[ \begin{matrix} -7 & 0 \\ 0 & -7 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 0 & -7 \\ -7 & 0 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 0 & 7 \\ 7 & 0 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Given \[A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]\] \[\therefore \] \[f(A)=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]-5\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]-\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -7 & 0 \\ 0 & -7 \\ \end{matrix} \right]\]
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