Here's a detailed, step-by-step solution to the problem, adhering to the requested structure and aiming for clarity and educational value.

1. Key Concepts and Formulas

   • Matrix Multiplication: The product of two matrices $A$ and $B$, denoted $AB$, is defined if the number of columns in $A$ equals the number of rows in $B$. If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, their product $C = AB$ is an $m \times p$ matrix where each element $c_{ij}$ is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
   • Powers of Matrices: For a square matrix $A$, $A^n$ is obtained by multiplying $A$ by itself $n$ times. For even powers, it's often more efficient to compute $A^n = (A^{n/2})^2$. For example, $A^4 = (A^2)^2$.
   • Solving Quadratic Equations: An equation of the form $ay^2 + by + c = 0$ can be solved by factoring, completing the square, or using the quadratic formula. When the variable is $x^2$ (e.g., $y=x^2$), remember that for real $x$, $x^2$ must be non-negative.

2. Step-by-Step Solution

   Step 1: Calculate $A^2$

   We are given the matrix $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$. To find $A^4$, it's computationally efficient to first calculate $A^2$ and then square the result. To find $A^2$, we multiply $A$ by itself: $A^2 = A \cdot A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ Let's perform the matrix multiplication:

   • The element in the first row, first column of $A^2$ is $(x)(x) + (1)(1) = x^2 + 1$.
   • The element in the first row, second column of $A^2$ is $(x)(1) + (1)(0) = x$.
   • The element in the second row, first column of $A^2$ is $(1)(x) + (0)(1) = x$.
   • The element in the second row, second column of $A^2$ is $(1)(1) + (0)(0) = 1$.

   Therefore, the matrix $A^2$ is: $A^2 = \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix}$

   Step 2: Calculate $A^4$ by computing $(A^2)^2$

   Now we compute $A^4$ by multiplying $A^2$ by itself: $A^4 = (A^2)^2 = \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix} \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix}$ Let's perform this matrix multiplication:

   • The element $(A^4)_{11}$ (first row, first column) is the dot product of the first row of $A^2$ and the first column of $A^2$: $(x^2+1)(x^2+1) + (x)(x) = (x^2+1)^2 + x^2$ Expanding this expression: $(x^4 + 2x^2 + 1) + x^2 = x^4 + 3x^2 + 1$.
   • The element $(A^4)_{12}$ (first row, second column) is: $(x^2+1)(x) + (x)(1) = x^3 + x + x = x^3 + 2x$.
   • The element $(A^4)_{21}$ (second row, first column) is: $(x)(x^2+1) + (1)(x) = x^3 + x + x = x^3 + 2x$.
   • The element $(A^4)_{22}$ (second row, second column) is: $(x)(x) + (1)(1) = x^2 + 1$.

   So, the matrix $A^4 = [a_{ij}]$ is: $A^4 = \begin{bmatrix} x^4 + 3x^2 + 1 & x^3 + 2x \\ x^3 + 2x & x^2 + 1 \end{bmatrix}$

   Step 3: Use the given condition $a_{11} = 109$ to find $x^2$

   We are given that $a_{11} = 109$. From our calculation in Step 2, the element $a_{11}$ of $A^4$ is $x^4 + 3x^2 + 1$. Therefore, we set up the equation: $x^4 + 3x^2 + 1 = 109$ Rearrange the equation into a standard quadratic form by setting $y = x^2$. Since $x \in \mathbb{R}$, $x^2$ must be non-negative, so $y \ge 0$. $x^4 + 3x^2 - 108 = 0$ $y^2 + 3y - 108 = 0$ We can solve this quadratic equation for $y$ by factoring. We look for two numbers that multiply to -108 and add to 3. These numbers are 12 and -9. $(y + 12)(y - 9) = 0$ This gives two possible values for $y$: $y = -12$ or $y = 9$. Since $y = x^2$ must be non-negative for real $x$, we discard $y = -12$. Thus, we must have $y = 9$. Substituting back $x^2 = y$: $x^2 = 9$

   Step 4: Calculate $a_{22}$

   We need to find the value of $a_{22}$, which is the element in the second row, second column of $A^4$. From our calculation in Step 2, $a_{22} = x^2 + 1$. From Step 3, we found that $x^2 = 9$. Substitute this value into the expression for $a_{22}$: $a_{22} = (9) + 1$ $a_{22} = 10$

3. Common Mistakes & Tips

   • Matrix Multiplication Errors: The most frequent mistake is incorrect calculation of dot products during matrix multiplication. Always double-check each element.
   • Efficiency for Powers: For $A^4$, calculating $A^2$ first and then $(A^2)^2$ is generally less error-prone than $A \cdot A \cdot A \cdot A$.
   • Quadratic in $x^2$: When solving equations like $x^4 + Px^2 + Q = 0$, remember to substitute $y=x^2$ and ensure $y \ge 0$ for real solutions of $x$.

4. Summary

   We first computed $A^2$ by multiplying matrix $A$ by itself. Then, we calculated $A^4$ by squaring $A^2$. This provided us with the expressions for $a_{11}$ and $a_{22}$ in terms of $x$. Using the given condition $a_{11} = 109$, we formed a quadratic equation in $x^2$ and solved it, finding $x^2=9$. Finally, we substituted this value of $x^2$ into the expression for $a_{22}$ to determine its value.

The final answer is $\boxed{10}$.