A) \[-8\]
B) \[-2\]
C) 0
D) \[1\]
E) \[8\]
Correct Answer: B
Solution :
\[A=\left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right]\] and \[{{A}^{2}}=\]Unit Matrix (given) ... (i) \[{{A}^{2}}=\left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} {{x}^{2}}+1 & x \\ x & 1 \\ \end{matrix} \right]\] ?. (ii) \[\because \] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] [From Eq.(i)] ...(iii) From Eqs. (ii) and (iii), \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} {{x}^{2}}+1 & x \\ x & 1 \\ \end{matrix} \right]\] On comparing, \[x=0\] Then, \[{{x}^{3}}+x-2=-2\]
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