Let $A=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array} ight], x \in R$ and $A^{4}=\left[a_{i j} i

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Let $A=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight], x \in R$ and $A^{4}=\left[a_{i j}ight]$. If $a_{11}=109$, then $a_{22}$ is equal to

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(10) : We have, $A=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]$\ $	herefore \quad A^{2}=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]=\left[\begin{array}{cc}x^{2}+1 & x \ x & 1\end{array}ight]$\ So, $\quad A^{4}=\left[\begin{array}{cc}x^{4}+1+3 x^{2} & x^{3}+2 x \ x^{3}+2 x & x^{2}+1\end{array}ight]$\ Now, $a_{11}=109 \Rightarrow x^{4}+3 x^{2}+1=109$\ $\Rightarrow x^{4}+12 x^{2}-9 x^{2}-108=0$\ $\Rightarrow \quad\left(x^{2}+12ight)\left(x^{2}-9ight)=0$\ $\Rightarrow \quad x^{2}=-12$ (not possible)\ $	herefore \quad x^{2}=9 \Rightarrow x= \pm 3$\ So, $\quad a_{22}=x^{2}+1=9+1=10$\