If `alpha and beta` are the roots of the equation, `[1,5][(1,3),(-4,7)]^(2)[(7/19, -13/19),(4/19,1/19)]^(4)`
`[(1,3),(-4,7)]^(2)[(x^2 - 5x + 5),(-3)] = [-4]`, then the value of `(2 - alpha)(2 - beta)` is

Options

A. 51
B. `-12`
C. `12`
D. `-7`

Answer

To solve the problem step by step, we will first rewrite the equation and then find the roots, followed by calculating the desired expression.

### Step 1: Rewrite the equation
The equation given is:
\[ [(1, 5)][(1, 3), (-4, 7)]^2 \left[(\frac{7}{19}, -\frac{13}{19}), (\frac{4}{19}, \frac{1}{19})\right]^4 \left[(1, 3), (-4, 7)]^2 \left[(x^2 - 5x + 5), -3\right] = -4 \]

### Step 2: Simplify the matrix expressions
We need to compute the determinants and adjoints of the matrices involved. For the matrix \( A = \begin{pmatrix} 1 & 3 \\ -4 & 7 \end{pmatrix} \):
- The determinant \( |A| = (1)(7) - (3)(-4) = 7 + 12 = 19 \).
- The adjoint of \( A \) is \( \text{adj}(A) = \begin{pmatrix} 7 & -3 \\ 4 & 1 \end{pmatrix} \).

### Step 3: Find \( A^{-1} \)
Using the formula for the inverse of a 2x2 matrix:
\[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{19} \begin{pmatrix} 7 & -3 \\ 4 & 1 \end{pmatrix} \]

### Step 4: Substitute back into the equation
Now substitute \( A \) and \( A^{-1} \) back into the equation. The left-hand side simplifies to:
\[ \frac{1}{19} \begin{pmatrix} 7 & -3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -4 & 7 \end{pmatrix}^2 \left[(x^2 - 5x + 5), -3\right] = -4 \]

### Step 5: Solve for \( x \)
After simplifying and rearranging the equation, we arrive at:
\[ x^2 - 5x - 6 = 0 \]
This factors to:
\[ (x - 6)(x + 1) = 0 \]
Thus, the roots are \( x = 6 \) and \( x = -1 \).

### Step 6: Identify \( \alpha \) and \( \beta \)
Let \( \alpha = 6 \) and \( \beta = -1 \).

### Step 7: Calculate \( (2 - \alpha)(2 - \beta) \)
Now we need to find:
\[ (2 - \alpha)(2 - \beta) = (2 - 6)(2 + 1) = (-4)(3) = -12 \]

### Final Answer
The value of \( (2 - \alpha)(2 - \beta) \) is \( -12 \).