A) I
B) 14 I
C) 0
D) None of these
Correct Answer: B
Solution :
\[{{A}^{2}}=A.A=\left[ \begin{matrix} 3 & -5 \\ -4 & 2 \\ \end{matrix} \right]\text{ }\left[ \begin{matrix} 3 & -5 \\ -4 & 2 \\ \end{matrix} \right]\] \[\Rightarrow \,{{A}^{2}}=\left[ \begin{matrix} 29 & -25 \\ -20 & 24 \\ \end{matrix} \right]\] and \[5A=\left[ \begin{matrix} 15 & -25 \\ -20 & 10 \\ \end{matrix} \right]\] \ \[{{A}^{2}}-5A=14\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=14I\].
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