If A =` [(3,1),(-1,2)]` and `l= [(1,0),(0,1)]` then the correct statement is

Options

A. `A^(2) + 5A -7l =O`
B. `-A^(2)+5A+7l=O`
C. `A^(2)-5A+7l=O`
D. `A^(2)+5A+7l=O`

Answer

To solve the problem, we need to analyze the given matrices and find the characteristic equation for matrix A. Then we will use the Cayley-Hamilton theorem to derive the correct statement.

### Step-by-Step Solution:

1. **Identify the matrices**:
Given:
\[ A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

2. **Set up the characteristic equation**:
We need to find the characteristic polynomial of matrix A. The characteristic polynomial is given by the determinant of \(A - \lambda I\):
\[ A - \lambda I = \begin{pmatrix} 3 - \lambda & 1 \\ -1 & 2 - \lambda \end{pmatrix} \]

3. **Calculate the determinant**:
The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated as \(ad - bc\). Thus, we have:
\[ \text{det}(A - \lambda I) = (3 - \lambda)(2 - \lambda) - (1)(-1) \]
Simplifying this:
\[ = (3 - \lambda)(2 - \lambda) + 1 \]
\[ = (6 - 3\lambda - 2\lambda + \lambda^2) + 1 \]
\[ = \lambda^2 - 5\lambda + 7 \]

4. **Set the characteristic polynomial to zero**:
The characteristic equation is:
\[ \lambda^2 - 5\lambda + 7 = 0 \]

5. **Apply the Cayley-Hamilton theorem**:
According to the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. Therefore, we replace \(\lambda\) with \(A\):
\[ A^2 - 5A + 7I = 0 \]

6. **Conclusion**:
The correct statement derived from the above steps is:
\[ A^2 - 5A + 7I = 0 \]