If A = [ 1 0 2; 0 2 1; 2 0 3 ], find f(A), where f(x) = x^3 - 6x^2 + 7x + 3. Prove that: | x a a; a x a; a a x | = (x - a)^2(x + 2a).

Steps to Solve

1. Identify the Matrix A and Function f(x)

The given matrix is

$$ A = \begin{pmatrix} 0 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 \end{pmatrix} $$

The function is

$$ f(x) = x^3 - 6x^2 + 7x + 3 $$

2. Compute f(A)

To compute ( f(A) ), we need to evaluate ( A^2 ) and ( A^3 ).

First, calculate ( A^2 ):

$$ A^2 = A \cdot A = \begin{pmatrix} 0 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 6 \ 2 & 4 & 5 \ 6 & 6 & 11 \end{pmatrix} $$

Next, calculate ( A^3 ):

$$ A^3 = A^2 \cdot A = \begin{pmatrix} 4 & 0 & 6 \ 2 & 4 & 5 \ 6 & 6 & 11 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 12 & 24 & 30 \ 10 & 14 & 24 \ 24 & 30 & 51 \end{pmatrix} $$

3. Substitute into f(A)

Now substitute into ( f(A) ):

$$ f(A) = A^3 - 6A^2 + 7A + 3I $$

Where ( I ) is the identity matrix.

4. Calculate the Final Expression

Calculate each term separately:

$$ 6A^2 = 6 \begin{pmatrix} 4 & 0 & 6 \ 2 & 4 & 5 \ 6 & 6 & 11 \end{pmatrix} = \begin{pmatrix} 24 & 0 & 36 \ 12 & 24 & 30 \ 36 & 36 & 66 \end{pmatrix} $$

$$ 7A = 7 \begin{pmatrix} 0 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 14 \ 0 & 14 & 7 \ 14 & 0 & 21 \end{pmatrix} $$

$$ 3I = 3 \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{pmatrix} $$

Combine all results to get ( f(A) ):

$$ f(A) = A^3 - 6A^2 + 7A + 3I $$

Finally, calculate ( f(A) ).