A) \[\left[ \begin{matrix} 4 & -6 \\ 6 & 4 \\ \end{matrix} \right]\]
B) \[0\]
C) \[\left[ \begin{matrix} -6 & 2 \\ -2 & 6 \\ \end{matrix} \right]\]
D) \[5l\]
Correct Answer: A
Solution :
Given, \[A=\left[ \begin{matrix} 2 & -1 \\ 1 & 2 \\ \end{matrix} \right]\] Now,\[{{A}^{2}}=\left[ \begin{matrix} 2 & -1 \\ 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -1 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 4-1 & -2-2 \\ 2+2 & -1+4 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{A}^{2}}=\left[ \begin{matrix} 3 & -4 \\ 4 & 3 \\ \end{matrix} \right]\] Now, \[{{A}^{2}}+2A-3l=\left[ \begin{matrix} 3 & -4 \\ 4 & 3 \\ \end{matrix} \right]+2\left[ \begin{matrix} 2 & -1 \\ 1 & 2 \\ \end{matrix} \right]-3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 3+4-3-4-2-0 \\ 4+2-0\,\,3+4-3 \\ \end{matrix} \right]\,=\,\left[ \begin{matrix} 4 & -6 \\ 6 & 4 \\ \end{matrix} \right]\]
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