`if A =[{:(1,0,2),(0,2,1),(2,0,3):}] ,` prove that `A^3-6A^2+7A+2I=0`

Explanation

To prove that \( A^3 - 6A^2 + 7A + 2I = 0 \) for the matrix

\[ A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix}, \]

we will follow these steps:

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

To find \( A^2 \), we multiply \( A \) by itself:

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

Calculating this product:

- First row:
- \( 1 \cdot 1 + 0 \cdot 0 + 2 \cdot 2 = 1 + 0 + 4 = 5 \)
- \( 1 \cdot 0 + 0 \cdot 2 + 2 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 1 \cdot 2 + 0 \cdot 1 + 2 \cdot 3 = 2 + 0 + 6 = 8 \)

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

- Third row:
- \( 2 \cdot 1 + 0 \cdot 0 + 3 \cdot 2 = 2 + 0 + 6 = 8 \)
- \( 2 \cdot 0 + 0 \cdot 2 + 3 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 2 \cdot 2 + 0 \cdot 1 + 3 \cdot 3 = 4 + 0 + 9 = 13 \)

Thus,

\[ A^2 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix}. \]

### Step 2: Calculate \( A^3 \)

Now, we calculate \( A^3 = A^2 \cdot A \):

\[ A^3 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix}. \]

Calculating this product:

- First row:
- \( 5 \cdot 1 + 0 \cdot 0 + 8 \cdot 2 = 5 + 0 + 16 = 21 \)
- \( 5 \cdot 0 + 0 \cdot 2 + 8 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 5 \cdot 2 + 0 \cdot 1 + 8 \cdot 3 = 10 + 0 + 24 = 34 \)

- Second row:
- \( 2 \cdot 1 + 4 \cdot 0 + 5 \cdot 2 = 2 + 0 + 10 = 12 \)
- \( 2 \cdot 0 + 4 \cdot 2 + 5 \cdot 0 = 0 + 8 + 0 = 8 \)
- \( 2 \cdot 2 + 4 \cdot 1 + 5 \cdot 3 = 4 + 4 + 15 = 23 \)

- Third row:
- \( 8 \cdot 1 + 0 \cdot 0 + 13 \cdot 2 = 8 + 0 + 26 = 34 \)
- \( 8 \cdot 0 + 0 \cdot 2 + 13 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 8 \cdot 2 + 0 \cdot 1 + 13 \cdot 3 = 16 + 0 + 39 = 55 \)

Thus,

\[ A^3 = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix}. \]

### Step 3: Calculate \( 6A^2 \)

Now, we calculate \( 6A^2 \):

\[ 6A^2 = 6 \cdot \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} = \begin{pmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{pmatrix}. \]

### Step 4: Calculate \( 7A \)

Next, we calculate \( 7A \):

\[ 7A = 7 \cdot \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix}. \]

### Step 5: Calculate \( 2I \)

The identity matrix \( I \) of order 3 is:

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

Thus,

\[ 2I = 2 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. \]

### Step 6: Combine all results

Now, we substitute these results into the equation \( A^3 - 6A^2 + 7A + 2I \):

\[ A^3 - 6A^2 + 7A + 2I = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} - \begin{pmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{pmatrix} + \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix} + \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. \]

Calculating this step by step:

- First row:
- \( 21 - 30 + 7 + 2 = 0 \)
- \( 0 - 0 + 0 + 0 = 0 \)
- \( 34 - 48 + 14 + 0 = 0 \)

- Second row:
- \( 12 - 12 + 0 + 0 = 0 \)
- \( 8 - 24 + 14 + 2 = 0 \)
- \( 23 - 30 + 7 + 0 = 0 \)

- Third row:
- \( 34 - 48 + 14 + 0 = 0 \)
- \( 0 - 0 + 0 + 0 = 0 \)
- \( 55 - 78 + 21 + 2 = 0 \)

Thus, we have:

\[ A^3 - 6A^2 + 7A + 2I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0. \]

### Conclusion

Therefore, we have proved that

\[ A^3 - 6A^2 + 7A + 2I = 0. \]