If `f(x) = x^(2) - 2x - 3` then find f(A) when `A = [(1,2),(2,1)]`

Answer

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

### Step-by-Step Solution:

1. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \]
To perform the multiplication, we calculate each element:
- First row, first column: \( 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4 \)
- Second row, first column: \( 2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4 \)
- Second row, second column: \( 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \)

Thus,
\[ A^2 = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} \]

**Hint**: To multiply two matrices, use the dot product of rows and columns.

2. **Calculate \( 2A \)**:
\[ 2A = 2 \cdot \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 2 \\ 2 \cdot 2 & 2 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix} \]

**Hint**: To scale a matrix, multiply each element by the scalar.

3. **Calculate \( 3I \)** where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus,
\[ 3I = 3 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

**Hint**: The identity matrix has 1's on the diagonal and 0's elsewhere.

4. **Substitute into the function \( f(A) \)**:
\[ f(A) = A^2 - 2A - 3I \]
Substituting the values we calculated:
\[ f(A) = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

5. **Perform the matrix subtraction**:
- First, subtract \( 2A \) from \( A^2 \):
\[ \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 5 - 2 & 4 - 4 \\ 4 - 4 & 5 - 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]
- Now subtract \( 3I \):
\[ \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 3 - 3 & 0 - 0 \\ 0 - 0 & 3 - 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

6. **Final Result**:
\[ f(A) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Conclusion:
The final result is:
\[ f(A) = 0 \]