Let `{:A=[(1,2),(-5,1)]and A^(-1)=xA+yI:}`, then the values of x and y are

Options

A. `x=-1/11,y=2/11`
B. `x=-1/11,y=-2/11`
C. `x=1/11,y=2/11`
D. `x=1/11,y=-2/11`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) such that \( A^{-1} = xA + yI \), where \( A = \begin{pmatrix} 1 & 2 \\ -5 & 1 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Find the Characteristic Polynomial of Matrix \( A \)**:
The characteristic polynomial is given by \( \det(A - \lambda I) = 0 \).
\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 2 \\ -5 & 1 - \lambda \end{pmatrix} \]
The determinant is calculated as follows:
\[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (-5)(2) = (1 - \lambda)^2 + 10 \]
Expanding this gives:
\[ (1 - \lambda)^2 + 10 = \lambda^2 - 2\lambda + 11 \]

2. **Apply Cayley-Hamilton Theorem**:
According to the Cayley-Hamilton theorem, a matrix satisfies its own characteristic equation:
\[ A^2 - 2A + 11I = 0 \]
Rearranging this gives:
\[ A^2 = 2A - 11I \]

3. **Multiply by \( A^{-1} \)**:
To express \( A^{-1} \), we can multiply the entire equation by \( A^{-1} \):
\[ A^{-1}A^2 = A^{-1}(2A - 11I) \]
This simplifies to:
\[ A = 2I - 11A^{-1} \]
Rearranging gives:
\[ 11A^{-1} = 2I - A \]
Thus,
\[ A^{-1} = \frac{2I - A}{11} \]

4. **Express \( A^{-1} \) in terms of \( A \) and \( I \)**:
We can rewrite \( A^{-1} \):
\[ A^{-1} = -\frac{1}{11}A + \frac{2}{11}I \]

5. **Identify \( x \) and \( y \)**:
From the expression \( A^{-1} = xA + yI \), we can compare coefficients:
- \( x = -\frac{1}{11} \)
- \( y = \frac{2}{11} \)

### Final Answer:
The values of \( x \) and \( y \) are:
\[ x = -\frac{1}{11}, \quad y = \frac{2}{11} \]