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

Hint: In this question, we have to find the value of ${a_{22}}$ by using the given matrix $A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$. Firstly, we will find the value of ${A^4}$. We can rewrite the ${A^4}$ as ${A^2}{A^2}$ and find the value of ${A^2}$ then the result obtained will be multiplied twice to arrive the value of ${A^4}$. Then compare the values of ${a_{11}}$ and ${a_{22}}$ with ${A^4}$.

Formula Used:
The basic matrix multiplication formula are given as
1. $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix}\times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix}= \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$

Complete step by step solution:
Given that the matrix is$A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$
Firstly, we will find the value of ${A^4}$.
We try to rewrite the expression ${A^4}$ in such a way that the simplification becomes easier.
So, ${A^4}$ can be rewritten as ${A^2}{A^2}$.
Now, we will find the value of ${A^2}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^2} = \begin{bmatrix}x&1\\1&0\end{bmatrix} \times \begin{bmatrix}x&1\\1&0\end{bmatrix} \\ {A^2} = \begin{bmatrix}{x\left( x \right) + 1\left( 1 \right)}&{x\left( 1 \right) + 1\left( 0 \right)}\\{1\left( x \right) + 0\left( 1 \right)}&{1\left( 1 \right) + 0\left( 0 \right)}\end{bmatrix}$
Further, we will simplify the above matrix, we get
${A^2} = \begin{bmatrix}{{x^2} + 1}&{x + 0}\\{x + 0}&{1 + 0}\end{bmatrix} \\ {A^2} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}$
Furthermore, we will find the value of ${A^4}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^4} = {A^2}{A^2}\\{A^4} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix} \times \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}\\{A^4} = \begin{bmatrix}{\left( {{x^2} + 1} \right)\left( {{x^2} + 1} \right) + x\left( x \right)}&{\left( {{x^2} + 1} \right)x + x\left( 1 \right)}\\{x\left( {{x^2} + 1} \right) + 1\left( x \right)}&{x\left( x \right) + 1\left( 1 \right)}\end{bmatrix}$
Now, we will simplify the above matrix, we get
${A^4} = \begin{bmatrix}{{x^4} + {x^2} + {x^2} + 1 + {x^2}}&{{x^3} + x + x}\\{{x^3} + x + x}&{{x^2} + 1}\end{bmatrix}\\{A^4} = \begin{bmatrix}{{x^4}3 + {x^2} + 1}&{{x^3} + 2x}\\{{x^3} + 2x}&{{x^2} + 1}\end{bmatrix}$
As we know that the general terms of the matrix can be written as $A = \begin{bmatrix}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{bmatrix}$ and we are given that ${a_{11}} = 109$.
So, we will compare the value of ${a_{11}}$ with matrix ${A^4}$, we get
${a_{11}} = {x^4} + 3{x^2} + 1\\109 = {x^4} + 3{x^2} + 1$
Now, we will simplify the above equation using factorization method, we get
${x^4} + 3{x^2} - 108 = 0\\{x^4} + 12{x^2} - 9{x^2} - 108 = 0\\{x^2}\left( {{x^2} + 12} \right) - 9\left( {{x^2} + 12} \right) = 0$
Further, we will factor out the common factor, we get
$\left( {{x^2} + 12} \right)\left( {{x^2} - 9} \right) = 0\\{x^2} = - 12,9$
As it is given that $x \in R$ so ${x^2} = - 12$ cannot be possible because the value under root cannot be negative.
Thus, $x = \pm 3$.
Now, we will find the value of ${a_{22}}$ by comparing it with matrix ${A^4}$, we get
${a_{22}} = {x^2} + 1$
Further, substitute ${x^2} = 9$, we get
${a_{22}} = 9 + 1\\{a_{22}} = 10$

Hence, the value of ${a_{22}}$ is $10$.

Note: Try to reduce the given terms so that we can solve them easily. Otherwise find the terms like ${A^2},{A^3}$ and ${A^4}$ to reduce the chances of mistakes.