In the matrix `A=[[x,1],[1,0]]` and `A^4=[[109,a_(12)],[a_(21),a_(22)]]`, then find the value of `a_(22)` is equal to

Options

A. 9
B. 10
C. 11
D. 12

Answer

To solve the problem, we need to compute the fourth power of the matrix \( A = \begin{pmatrix} x & 1 \\ 1 & 0 \end{pmatrix} \) and find the value of \( a_{22} \) in the resulting matrix \( A^4 = \begin{pmatrix} 109 & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)

First, we compute \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} x & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} x & 1 \\ 1 & 0 \end{pmatrix} \]

Calculating the entries:

- The (1,1) entry: \( x \cdot x + 1 \cdot 1 = x^2 + 1 \)
- The (1,2) entry: \( x \cdot 1 + 1 \cdot 0 = x \)
- The (2,1) entry: \( 1 \cdot x + 0 \cdot 1 = x \)
- The (2,2) entry: \( 1 \cdot 1 + 0 \cdot 0 = 1 \)

Thus, we have:

\[ A^2 = \begin{pmatrix} x^2 + 1 & x \\ x & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^4 \)

Next, we compute \( A^4 = A^2 \cdot A^2 \):

\[ A^4 = \begin{pmatrix} x^2 + 1 & x \\ x & 1 \end{pmatrix} \cdot \begin{pmatrix} x^2 + 1 & x \\ x & 1 \end{pmatrix} \]

Calculating the entries:

- The (1,1) entry: \( (x^2 + 1)(x^2 + 1) + x \cdot x = (x^2 + 1)^2 + x^2 \)
- The (1,2) entry: \( (x^2 + 1)x + x \cdot 1 = x^3 + x + x = x^3 + 2x \)
- The (2,1) entry: \( x(x^2 + 1) + 1 \cdot x = x^3 + x + x = x^3 + 2x \)
- The (2,2) entry: \( x \cdot x + 1 \cdot 1 = x^2 + 1 \)

Thus, we have:

\[ A^4 = \begin{pmatrix} (x^2 + 1)^2 + x^2 & x^3 + 2x \\ x^3 + 2x & x^2 + 1 \end{pmatrix} \]

### Step 3: Set up the equations

We know that:

\[ A^4 = \begin{pmatrix} 109 & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]

From this, we can equate:

1. \( (x^2 + 1)^2 + x^2 = 109 \)
2. \( a_{12} = x^3 + 2x \)
3. \( a_{21} = x^3 + 2x \)
4. \( a_{22} = x^2 + 1 \)

### Step 4: Solve for \( x \)

We need to solve the equation:

\[ (x^2 + 1)^2 + x^2 = 109 \]

Let \( y = x^2 \). Then we have:

\[ (y + 1)^2 + y = 109 \]

Expanding this gives:

\[ y^2 + 2y + 1 + y = 109 \implies y^2 + 3y + 1 - 109 = 0 \implies y^2 + 3y - 108 = 0 \]

### Step 5: Factor the quadratic equation

We can factor this quadratic:

\[ (y + 12)(y - 9) = 0 \]

Thus, \( y = -12 \) or \( y = 9 \). Since \( y = x^2 \), we discard \( y = -12 \) and take \( y = 9 \):

\[ x^2 = 9 \implies x = 3 \quad (\text{since } x \text{ is real}) \]

### Step 6: Find \( a_{22} \)

Now, we can find \( a_{22} \):

\[ a_{22} = x^2 + 1 = 9 + 1 = 10 \]

### Final Answer

Thus, the value of \( a_{22} \) is:

\[ \boxed{10} \]