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

Answer

To solve the problem, we need to show that \( A^2 - 5A + 7I = O \) for the matrix \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \), and then find the inverse of \( A \).

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]

Calculating the elements of the resulting matrix:

- First row, first column:
\[ 3 \cdot 3 + 1 \cdot (-1) = 9 - 1 = 8 \]

- First row, second column:
\[ 3 \cdot 1 + 1 \cdot 2 = 3 + 2 = 5 \]

- Second row, first column:
\[ -1 \cdot 3 + 2 \cdot (-1) = -3 - 2 = -5 \]

- Second row, second column:
\[ -1 \cdot 1 + 2 \cdot 2 = -1 + 4 = 3 \]

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

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

Now, we calculate \( 5A \):

\[ 5A = 5 \cdot \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} \]

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

The identity matrix \( I \) for a 2x2 matrix is:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus,
\[ 7I = 7 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]

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

Now we substitute \( A^2 \), \( 5A \), and \( 7I \) into the equation:

\[ A^2 - 5A + 7I = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \]

Calculating each element:

- First row, first column:
\[ 8 - 15 + 7 = 0 \]

- First row, second column:
\[ 5 - 5 + 0 = 0 \]

- Second row, first column:
\[ -5 + 5 + 0 = 0 \]

- Second row, second column:
\[ 3 - 10 + 7 = 0 \]

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

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

To find the inverse of \( A \), we use the formula for the inverse of a 2x2 matrix:

\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

Where \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).

For our matrix \( A \):
- \( a = 3 \)
- \( b = 1 \)
- \( c = -1 \)
- \( d = 2 \)

Calculating the determinant \( \text{det}(A) \):

\[ \text{det}(A) = ad - bc = (3)(2) - (1)(-1) = 6 + 1 = 7 \]

Now substituting into the inverse formula:

\[ A^{-1} = \frac{1}{7} \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{2}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{3}{7} \end{pmatrix} \]

### Final Result

Thus, we have shown that:

\[ A^2 - 5A + 7I = O \]

And the inverse of \( A \) is:

\[ A^{-1} = \begin{pmatrix} \frac{2}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{3}{7} \end{pmatrix} \]