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

Explanation

To find \( f(A) \) where \( f(x) = x^2 - 5x + 6 \) and \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
We need to compute the square of matrix \( A \).

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

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

- First row:
- \( (2 \cdot 2 + 0 \cdot 2 + 1 \cdot 1) = 4 + 0 + 1 = 5 \)
- \( (2 \cdot 0 + 0 \cdot 1 + 1 \cdot -1) = 0 + 0 - 1 = -1 \)
- \( (2 \cdot 1 + 0 \cdot 3 + 1 \cdot 0) = 2 + 0 + 0 = 2 \)

- Second row:
- \( (2 \cdot 2 + 1 \cdot 2 + 3 \cdot 1) = 4 + 2 + 3 = 9 \)
- \( (2 \cdot 0 + 1 \cdot 1 + 3 \cdot -1) = 0 + 1 - 3 = -2 \)
- \( (2 \cdot 1 + 1 \cdot 3 + 3 \cdot 0) = 2 + 3 + 0 = 5 \)

- Third row:
- \( (1 \cdot 2 + -1 \cdot 2 + 0 \cdot 1) = 2 - 2 + 0 = 0 \)
- \( (1 \cdot 0 + -1 \cdot 1 + 0 \cdot -1) = 0 - 1 + 0 = -1 \)
- \( (1 \cdot 1 + -1 \cdot 3 + 0 \cdot 0) = 1 - 3 + 0 = -2 \)

Thus,

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

### Step 2: Calculate \( -5A \)
Now we compute \( -5A \):

\[ -5A = -5 \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix} \]

### Step 3: Calculate \( 6I \)
Next, we compute \( 6I \), where \( I \) is the identity matrix of size 3:

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

Thus,

\[ 6I = 6 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

### Step 4: Combine to find \( f(A) \)
Now we can find \( f(A) \) using the formula:

\[ f(A) = A^2 - 5A + 6I \]

Calculating \( f(A) \):

\[ f(A) = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} + \begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix} + \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

Now we add the matrices element-wise:

- 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,

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

### Final Result
The final result is:

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