A) A
B) \[{{A}^{2}}\]
C) \[{{A}^{3}}\]
D) \[{{A}^{4}}\]
Correct Answer: C
Solution :
Here, \[=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{A}^{2}}=A.A=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\,\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right].\] \[=-{{\sin }^{2}}A+\sin A\sin (B+C)\] Þ \[+\cos B(\cos B+\cos A\cos C)\] Þ \[(x-5)({{x}^{2}}+5x-14)=0\] \[{{x}^{3}}-39x+70=0\] Now, \[{{A}^{2}}=\left[ \,\begin{matrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \\ \end{matrix}\, \right]\] and \[{{A}^{3}}={{A}^{2}}.A=\left[ \begin{matrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \\ \end{matrix} \right]\,\times \,\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \\ \end{matrix} \right]\] \[=\left[ \,\begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \\ \end{matrix}\, \right]={{A}^{-1}}\].
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