If `A=[[3, 1], [-1, 2]] and A^(2)-5A+7I=0`, then `I=`

Options

A. `(1)/(5)(A)+(7)/(5)(A^(-1))`
B. `(1)/(7)(A)+(5)/(7)(A^(-1))`
C. `(1)/(7)(A)-(5)/(7)(A^(-1))`
D. `(1)/(5)(A)-(7)/(5)(A^(-1))`

Answer

To solve the equation \( A^2 - 5A + 7I = 0 \) for \( I \), we will follow these steps:

1. **Write down the given matrix \( A \)**:
\[ A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]

2. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]
\[ = \begin{pmatrix} 3 \cdot 3 + 1 \cdot (-1) & 3 \cdot 1 + 1 \cdot 2 \\ -1 \cdot 3 + 2 \cdot (-1) & -1 \cdot 1 + 2 \cdot 2 \end{pmatrix} \]
\[ = \begin{pmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} \]

3. **Substitute \( A^2 \) into the equation \( A^2 - 5A + 7I = 0 \)**:
\[ \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - 5 \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} + 7I = 0 \]
\[ = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} + 7I = 0 \]

4. **Combine the matrices**:
\[ \begin{pmatrix} 8 - 15 & 5 - 5 \\ -5 + 5 & 3 - 10 \end{pmatrix} + 7I = 0 \]
\[ = \begin{pmatrix} -7 & 0 \\ 0 & -7 \end{pmatrix} + 7I = 0 \]

5. **Rearranging gives us**:
\[ 7I = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]

6. **Divide by 7**:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus, the identity matrix \( I \) is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]