Let A = begin{bmatrix} x & 1 1 & 0 end{bmatr | vedclass

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(A) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} \begin{bmat

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$10$

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(A) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^{2} + 1 & x \ x & 1 \end{bmatrix}$.Next,calculate $A^{4} = A^{2} 	imes A^{2}$:$A^{4} = \begin{bmatrix} x^{2} + 1 & x \ x & 1 \end{bmatrix} \begin{bmatrix} x^{2} + 1 & x \ x & 1 \end{bmatrix} = \begin{bmatrix} (x^{2} + 1)^{2} + x^{2} & x(x^{2} + 1) + x \ x(x^{2} + 1) + x & x^{2} + 1 \end{bmatrix}$.We are given $a_{11} = 109$,so:$(x^{2} + 1)^{2} + x^{2} = 109$.Let $y = x^{2}$. Then $(y + 1)^{2} + y = 109$.$y^{2} + 2y + 1 + y = 109 \Rightarrow y^{2} + 3y - 108 = 0$.$(y + 12)(y - 9) = 0$.Since $y = x^{2} \geq 0$,we have $y = 9$,so $x^{2} = 9$.Now,find $a_{22}$:From the matrix $A^{4}$,$a_{22} = x^{2} + 1$.Substituting $x^{2} = 9$,we get $a_{22} = 9 + 1 = 10$.

Let $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$,$x \in \mathbb{R}$ and $A^{4} = [a_{ij}]$. If $a_{11} = 109$,then $a_{22}$ is equal to

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