If `A[(1,0,2),(0,2,1),(2,0,3)]` and `A^(3)-6A^(2)+7A+kI_(3)=O` find k.

Answer

To solve the problem, we need to find the value of \( k \) in the equation \( A^3 - 6A^2 + 7A + kI_3 = O \), where \( A \) is given as:

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

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

- 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, we have:

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

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

Now, we calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):

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

Calculating the elements of \( A^3 \):

- 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, we have:

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

### Step 3: Substitute into the equation

Now, we substitute \( A^3 \), \( A^2 \), and \( A \) into the equation \( A^3 - 6A^2 + 7A + kI_3 = O \):

\[ \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} - 6 \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} + 7 \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} + k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

Calculating each term:

1. **Calculate \( -6A^2 \)**:
\[ -6A^2 = -6 \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} = \begin{pmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{pmatrix} \]

2. **Calculate \( 7A \)**:
\[ 7A = 7 \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix} \]

3. **Combine all terms**:
\[ A^3 - 6A^2 + 7A + kI_3 = \begin{pmatrix} 21 - 30 + 7 + k & 0 + 0 + 0 & 34 - 48 + 14 \\ 12 - 12 + 0 & 8 - 24 + 14 & 23 - 30 + 7 \\ 34 - 48 + 14 & 0 + 0 + 0 & 55 - 78 + 21 \end{pmatrix} \]

This simplifies to:
\[ = \begin{pmatrix} -2 + k & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 + k \end{pmatrix} \]

### Step 4: Set the matrix equal to zero

Setting the resulting matrix equal to the zero matrix gives us the equations:

1. \( -2 + k = 0 \)
2. \( -2 = 0 \) (which is not possible)
3. \( -2 + k = 0 \)

From the first and third equations, we find:

\[ k = 2 \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{2} \]