If `A=[(3,1),(-1,2)]`, then find f (A),
where `f(x)=x^(2)-5x+7`.

Answer

To solve the problem, we need to find \( f(A) \) where \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \) and \( f(x) = x^2 - 5x + 7 \).

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

To find \( f(A) \), we first need to compute \( A^2 \):

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

Calculating the elements of \( A^2 \):

- 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: Compute \( 5A \)

Next, we compute \( 5A \):

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

### Step 3: Compute \( f(A) \)

Now we can compute \( f(A) \):

\[ f(A) = A^2 - 5A + 7I \]

Where \( I \) is the identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), so \( 7I = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \).

Now substituting the values we calculated:

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

### Step 4: Perform the matrix subtraction and addition

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

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

Now add \( 7I \):

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

### Final Result

Thus, we find that:

\[ f(A) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]