If `A=[{:(2,-1),(1,3):}]` , then show that `A^(2)-5A+7I_(2)=O`, hence find `A^(-1)`.

Answer

To solve the problem, we need to show that \( A^2 - 5A + 7I = O \) and then find \( A^{-1} \).

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

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \]

Calculating the elements of \( A^2 \):
- First row, first column: \( 2 \cdot 2 + (-1) \cdot 1 = 4 - 1 = 3 \)
- First row, second column: \( 2 \cdot (-1) + (-1) \cdot 3 = -2 - 3 = -5 \)
- Second row, first column: \( 1 \cdot 2 + 3 \cdot 1 = 2 + 3 = 5 \)
- Second row, second column: \( 1 \cdot (-1) + 3 \cdot 3 = -1 + 9 = 8 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 3 & -5 \\ 5 & 8 \end{pmatrix} \]

### Step 2: Calculate \( 5A \)

Next, we calculate \( 5A \):
\[ 5A = 5 \cdot \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 10 & -5 \\ 5 & 15 \end{pmatrix} \]

### Step 3: Calculate \( 7I \)

Now, we calculate \( 7I \) where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \implies 7I = 7 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]

### Step 4: Calculate \( A^2 - 5A + 7I \)

Now we substitute \( A^2 \), \( 5A \), and \( 7I \) into the equation:
\[ A^2 - 5A + 7I = \begin{pmatrix} 3 & -5 \\ 5 & 8 \end{pmatrix} - \begin{pmatrix} 10 & -5 \\ 5 & 15 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]

Calculating the matrix:
- First row, first column: \( 3 - 10 + 7 = 0 \)
- First row, second column: \( -5 - (-5) + 0 = 0 \)
- Second row, first column: \( 5 - 5 + 0 = 0 \)
- Second row, second column: \( 8 - 15 + 7 = 0 \)

Thus, we have:
\[ A^2 - 5A + 7I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O \]

### Step 5: Find \( A^{-1} \)

To find \( A^{-1} \), we can use the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

First, we calculate the determinant of \( A \):
\[ \text{det}(A) = (2)(3) - (-1)(1) = 6 + 1 = 7 \]

Next, we find the adjugate of \( A \):
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]

Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{7} \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{7} & \frac{1}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} \]

### Final Answer

Thus, we have shown that \( A^2 - 5A + 7I = O \) and found:
\[ A^{-1} = \begin{pmatrix} \frac{3}{7} & \frac{1}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} \]