If `f(x)=x^2-5x+6.` Find `f(A),if A=[(2,0,1),(2,1,3),(1,-1,0)]`.

Options

A. `[{:(1,-1,-3),(-1,1,10),(-5,4,4):}]`
B. `[{:(1,-1,-3),(-1,-1,-10),(-5,-4,-4):}]`
C. `[{:(1,-1,-3),(-1,-1,-10),(-5,4,4):}]`
D. None of these

Explanation

To solve the problem, we need to evaluate the function \( f(A) \) where \( f(x) = x^2 - 5x + 6 \) and \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Identify the Function**:
We have the function \( f(x) = x^2 - 5x + 6 \).

2. **Substitute the Matrix**:
To find \( f(A) \), we substitute \( A \) into the function:
\[ f(A) = A^2 - 5A + 6I \]
where \( I \) is the identity matrix of the same order as \( A \) (which is \( 3 \times 3 \)).

3. **Calculate \( A^2 \)**:
We need to compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]
Performing the multiplication:
- First row:
- First column: \( 2 \cdot 2 + 0 \cdot 2 + 1 \cdot 1 = 4 + 0 + 1 = 5 \)
- Second column: \( 2 \cdot 0 + 0 \cdot 1 + 1 \cdot (-1) = 0 + 0 - 1 = -1 \)
- Third column: \( 2 \cdot 1 + 0 \cdot 3 + 1 \cdot 0 = 2 + 0 + 0 = 2 \)
- Second row:
- First column: \( 2 \cdot 2 + 1 \cdot 2 + 3 \cdot 1 = 4 + 2 + 3 = 9 \)
- Second column: \( 2 \cdot 0 + 1 \cdot 1 + 3 \cdot (-1) = 0 + 1 - 3 = -2 \)
- Third column: \( 2 \cdot 1 + 1 \cdot 3 + 3 \cdot 0 = 2 + 3 + 0 = 5 \)
- Third row:
- First column: \( 1 \cdot 2 + (-1) \cdot 2 + 0 \cdot 1 = 2 - 2 + 0 = 0 \)
- Second column: \( 1 \cdot 0 + (-1) \cdot 1 + 0 \cdot (-1) = 0 - 1 + 0 = -1 \)
- Third column: \( 1 \cdot 1 + (-1) \cdot 3 + 0 \cdot 0 = 1 - 3 + 0 = -2 \)

Thus,
\[ A^2 = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} \]

4. **Calculate \( 5A \)**:
Now we calculate \( 5A \):
\[ 5A = 5 \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{pmatrix} \]

5. **Calculate \( 6I \)**:
The identity matrix \( I \) for \( 3 \times 3 \) is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Therefore,
\[ 6I = 6 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

6. **Combine the Results**:
Now we substitute back into the equation:
\[ f(A) = A^2 - 5A + 6I \]
\[ f(A) = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} - \begin{pmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{pmatrix} + \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]
Performing the subtraction and addition:
- First row:
- \( 5 - 10 + 6 = 1 \)
- \( -1 - 0 + 0 = -1 \)
- \( 2 - 5 + 0 = -3 \)
- Second row:
- \( 9 - 10 + 0 = -1 \)
- \( -2 - 5 + 6 = -1 \)
- \( 5 - 15 + 0 = -10 \)
- Third row:
- \( 0 - 5 + 0 = -5 \)
- \( -1 + 5 + 0 = 4 \)
- \( -2 - 0 + 6 = 4 \)

Thus, we get:
\[ f(A) = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix} \]

### Final Answer:
\[ f(A) = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix} \]