A) 1
B) 2
C) 3
D) 0
Correct Answer: D
Solution :
\[A=\left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right],\therefore {{A}^{2}}=I\Rightarrow \left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right]\,\left[ \begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] Þ \[\left[ \begin{matrix} {{x}^{2}}+1 & x \\ x & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\Rightarrow {{x}^{2}}+1=1\Rightarrow x=0\].
You need to login to perform this action.
You will be redirected in
3 sec
