Let \(\mathrm{A}=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbb{R}^{+}\)and \(\mathrm{A}^4=\left[\mathrm{a}_{\mathrm{ij}}\right]_2\).
If \(\mathrm{a}_{11}=109\), then \(\left(\mathrm{A}^4\right)^{-1}=\)

\(\begin{aligned} & A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right] \\ & A^2=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}x^2+1 & x \\ x & 1\end{array}\right] \\ \therefore \quad & A^4=A^2 \cdot A^2\end{aligned}\)
\(\begin{aligned}
& =\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right]\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
\left(x^2+1\right)^2+x^2 & x\left(x^2+1+1\right) \\
x\left(x^2+1+1\right) & \left(x^2+1\right)
\end{array}\right] \\
& =\left[\begin{array}{cc}
\left(x^2+1\right)^2+x^2 & x\left(x^2+2\right) \\
x\left(x^2+2\right) & x^2+1
\end{array}\right] \\
\therefore \quad \begin{array}{l}
\dot{A}^4
\end{array} & =\left[\mathrm{a}_{\mathrm{ij}}\right] \text { and } \mathrm{a}_{11}=109 \\
\mathrm{a}_{11} & =109 \\
\Rightarrow & \Rightarrow\left(x^2+1\right)^2+x^2=109
\end{aligned}\)
...[Given]
i.e.
\(\begin{aligned}
\left(x^2+1\right)^2+x^2 & =100+9 \\
& =(10)^2+3^2=\left(3^2+1\right)^2+3^2
\end{aligned}\)
Comparing we get,
\(x^2=9 \Rightarrow x=3\)
\(\mathrm{a}_{12}=x\left(x^2+2\right)=3(9+2)=33\)
\(\begin{aligned}
& \mathrm{a}_{21}=x\left(x^2+2\right)=3(9+2)=33 \\
& \mathrm{a}_{22}=x^2+1=9+1=10
\end{aligned}\)
\(\begin{array}{ll}
\therefore & A^4=\left[\begin{array}{cc}
109 & 33 \\
33 & 10
\end{array}\right] \\
\therefore & \left|A^4\right|=1 \neq 0
\end{array}\)
If \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and \(a d-b c \neq 0\), then
\(\begin{array}{ll}
& A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
\therefore & \left(A^4\right)^{-1}=\left[\begin{array}{cc}
10 & -33 \\
-33 & 109
\end{array}\right]
\end{array}\)