A) \[\left[ \begin{matrix} 0 & -4 \\ 8 & 8 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 2 & 1 \\ 2 & 0 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 8 & 4 \\ 8 & 0 \\ \end{matrix} \right]\]
E) \[\left[ \begin{matrix} 1 & 0 \\ 1 & 2 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
\[\because \] \[A=\left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 9 & -4 \\ -8 & 17 \\ \end{matrix} \right]\] \[\therefore \] \[f(A)={{A}^{2}}+4A-5\] \[=\left[ \begin{matrix} 9 & -4 \\ -8 & 17 \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & 8 \\ 16 & -12 \\ \end{matrix} \right]-\left[ \begin{matrix} 5 & 0 \\ 0 & 5 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 9+4-5 & -4+8 \\ -8+16 & 17-12-5 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 4 \\ 8 & 0 \\ \end{matrix} \right]\]
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