Let A=left[begin{array}{ll}x & 1 1 & 0end{ar | vedclass

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(D) Given $A=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]$.First,calculate $A^2$:$A^2 = \left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}i

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$\left[\begin{array}{cc}10 & -33 \ -33 & 109\end{array}ight]$

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(D) Given $A=\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]$.First,calculate $A^2$:$A^2 = \left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight]\left[\begin{array}{ll}x & 1 \ 1 & 0\end{array}ight] = \left[\begin{array}{cc}x^2+1 & x \ x & 1\end{array}ight]$.Now,calculate $A^4 = A^2 \cdot A^2$:$A^4 = \left[\begin{array}{cc}x^2+1 & x \ x & 1\end{array}ight]\left[\begin{array}{cc}x^2+1 & x \ x & 1\end{array}ight] = \left[\begin{array}{cc}(x^2+1)^2+x^2 & x(x^2+1+1) \ x(x^2+1+1) & x^2+1\end{array}ight] = \left[\begin{array}{cc}(x^2+1)^2+x^2 & x(x^2+2) \ x(x^2+2) & x^2+1\end{array}ight]$.Given $a_{11} = 109$,we have $(x^2+1)^2 + x^2 = 109$.Let $t = x^2$. Then $(t+1)^2 + t = 109 \Rightarrow t^2 + 2t + 1 + t = 109 \Rightarrow t^2 + 3t - 108 = 0$.Solving for $t$: $(t+12)(t-9) = 0$. Since $x \in R^{+}$,$t = x^2 = 9 \Rightarrow x = 3$.Substituting $x=3$ into the matrix $A^4$:$a_{11} = (9+1)^2 + 9 = 109$,$a_{12} = a_{21} = 3(9+2) = 33$,$a_{22} = 9+1 = 10$.So,$A^4 = \left[\begin{array}{cc}109 & 33 \ 33 & 10\end{array}ight]$.The determinant $|A^4| = (109)(10) - (33)(33) = 1090 - 1089 = 1$.The inverse is given by $\frac{1}{|A^4|} 	ext{adj}(A^4) = \frac{1}{1} \left[\begin{array}{cc}10 & -33 \ -33 & 109\end{array}ight] = \left[\begin{array}{cc}10 & -33 \ -33 & 109\end{array}ight]$.

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

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