Suppose A is any 3 × 3 non-singular matrix and (A -3I) (A-5I)=0, where `I=I_3` and `O=O_3` . If `alphaA + betaA^(-1)` =4I , then `alpha +beta` is equal to :

Options

A. 13
B. 7
C. 12
D. 8

Answer

To solve the problem step by step, we will analyze the given equation and derive the necessary values.

### Step 1: Analyze the given equation
We are given that \((A - 3I)(A - 5I) = 0\). This implies that the eigenvalues of the matrix \(A\) are 3 and 5.

### Step 2: Expand the equation
We can expand the equation:
\[ (A - 3I)(A - 5I) = A^2 - 5A - 3A + 15I = A^2 - 8A + 15I = 0 \]
Thus, we can rearrange this to:
\[ A^2 - 8A + 15I = 0 \]
or
\[ A^2 = 8A - 15I \]

### Step 3: Pre-multiply by \(A^{-1}\)
Since \(A\) is non-singular, we can pre-multiply both sides of the equation by \(A^{-1}\):
\[ A^{-1}A^2 = A^{-1}(8A - 15I) \]
This simplifies to:
\[ A = 8I - 15A^{-1} \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation:
\[ A + 15A^{-1} = 8I \]

### Step 5: Compare with given equation
We are also given that \(\alpha A + \beta A^{-1} = 4I\). We can compare this with our derived equation:
\[ A + 15A^{-1} = 8I \]
This suggests that:
\[ \alpha = 1 \quad \text{and} \quad \beta = 15 \]

### Step 6: Calculate \(\alpha + \beta\)
Now, we can find \(\alpha + \beta\):
\[ \alpha + \beta = 1 + 15 = 16 \]

### Final Answer
Thus, the value of \(\alpha + \beta\) is:
\[ \boxed{16} \]