A) \[4\left[ \begin{matrix} 2 & 1 \\ 2 & 0 \\ \end{matrix} \right]\]
B) \[4\left[ \begin{matrix} 0 & -1 \\ 2 & 2 \\ \end{matrix} \right]\]
C) \[32\left[ \begin{matrix} 2 & 1 \\ 2 & 0 \\ \end{matrix} \right]\]
D) \[32\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right],\,{{A}^{2}}=A\cdot A=\left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 2 \\ 4 & -3 \\ \end{matrix} \right]\]\[=\left[ \begin{matrix} 9 & -4 \\ -8 & 7 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}+4A-5l=\left[ \begin{matrix} 9 & -4 \\ -8 & 17 \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & 8 \\ 16 & -12 \\ \end{matrix} \right]+\left[ \begin{matrix} -5 & 0 \\ 0 & -5 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 4 \\ 8 & 0 \\ \end{matrix} \right]=4\left[ \begin{matrix} 2 & 1 \\ 2 & 0 \\ \end{matrix} \right]\]
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