Let A = `[(3,-1),(0,2)]` . Suppose A satisfies the equation `x^(2)+ax+b=0` for some real numbers a and b . Let `alpha , beta ` be the roots of `t^(2)+at` +b=0 then
`(1)/(alpha^(2)-3alpha+4)+(1)/(beta^(2)-3alpha+4)`= ________

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript, while ensuring clarity in each step.

### Step 1: Identify the matrix A
Given the matrix:
\[ A = \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} \]

### Step 2: Find the characteristic polynomial
The characteristic polynomial of a matrix \( A \) is given by:
\[ \text{det}(A - \lambda I) = 0 \]
where \( I \) is the identity matrix.

Calculating \( A - \lambda I \):
\[ A - \lambda I = \begin{pmatrix} 3 - \lambda & -1 \\ 0 & 2 - \lambda \end{pmatrix} \]

The determinant is:
\[ \text{det}(A - \lambda I) = (3 - \lambda)(2 - \lambda) - (0)(-1) = (3 - \lambda)(2 - \lambda) \]

Setting the determinant to zero:
\[ (3 - \lambda)(2 - \lambda) = 0 \]
This gives us the eigenvalues:
\[ \lambda_1 = 3, \quad \lambda_2 = 2 \]

### Step 3: Form the quadratic equation
We know that the eigenvalues satisfy the equation:
\[ x^2 + ax + b = 0 \]
From the eigenvalues, we can find \( a \) and \( b \):
- The sum of the roots (eigenvalues) is \( 3 + 2 = 5 \), so \( a = -5 \).
- The product of the roots is \( 3 \cdot 2 = 6 \), so \( b = 6 \).

Thus, the quadratic equation is:
\[ t^2 - 5t + 6 = 0 \]

### Step 4: Find the roots
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting \( a = 1, b = -5, c = 6 \):
\[ t = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ t = \frac{5 \pm \sqrt{25 - 24}}{2} \]
\[ t = \frac{5 \pm 1}{2} \]
Thus, the roots are:
\[ \alpha = 3, \quad \beta = 2 \]

### Step 5: Evaluate the expression
We need to evaluate:
\[ \frac{1}{\alpha^2 - 3\alpha + 4} + \frac{1}{\beta^2 - 3\beta + 4} \]

Calculating for \( \alpha = 3 \):
\[ \alpha^2 - 3\alpha + 4 = 3^2 - 3 \cdot 3 + 4 = 9 - 9 + 4 = 4 \]
Thus,
\[ \frac{1}{\alpha^2 - 3\alpha + 4} = \frac{1}{4} \]

Calculating for \( \beta = 2 \):
\[ \beta^2 - 3\beta + 4 = 2^2 - 3 \cdot 2 + 4 = 4 - 6 + 4 = 2 \]
Thus,
\[ \frac{1}{\beta^2 - 3\beta + 4} = \frac{1}{2} \]

### Step 6: Combine the results
Now, we combine the two results:
\[ \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]

### Final Answer
The final answer is:
\[ \frac{3}{4} \]