`if A=[{:(3,1),(-1,2):}],`show that `A^(2)-5A+7I=0.`

Explanation

To solve the problem, we need to show that \( A^2 - 5A + 7I = 0 \) for the matrix \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \).

### 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:

- 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, we have:

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

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

Next, 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 \( 2 \times 2 \) matrix is:

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

Thus, we have:

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

### Step 4: Combine the results

Now we can substitute \( A^2 \), \( 5A \), and \( 7I \) into the expression \( A^2 - 5A + 7I \):

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

Calculating this step-by-step:

1. First, calculate \( A^2 - 5A \):

\[ \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} = \begin{pmatrix} 8 - 15 & 5 - 5 \\ -5 - (-5) & 3 - 10 \end{pmatrix} = \begin{pmatrix} -7 & 0 \\ 0 & -7 \end{pmatrix} \]

2. Now add \( 7I \):

\[ \begin{pmatrix} -7 & 0 \\ 0 & -7 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} = \begin{pmatrix} -7 + 7 & 0 + 0 \\ 0 + 0 & -7 + 7 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Conclusion

Thus, we have shown that:

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

This is the zero matrix, confirming that \( A^2 - 5A + 7I = 0 \).