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 entries of \( A^2 \):
- First row, first column: \( 1 \cdot 1 + 0 \cdot 0 + 2 \cdot 2 = 1 + 0 + 4 = 5 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 2 + 2 \cdot 0 = 0 + 0 + 0 = 0 \)
- First row, third column: \( 1 \cdot 2 + 0 \cdot 1 + 2 \cdot 3 = 2 + 0 + 6 = 8 \)
- Second row, first column: \( 0 \cdot 1 + 2 \cdot 0 + 1 \cdot 2 = 0 + 0 + 2 = 2 \)
- Second row, second column: \( 0 \cdot 0 + 2 \cdot 2 + 1 \cdot 0 = 0 + 4 + 0 = 4 \)
- Second row, third column: \( 0 \cdot 2 + 2 \cdot 1 + 1 \cdot 3 = 0 + 2 + 3 = 5 \)
- Third row, first column: \( 2 \cdot 1 + 0 \cdot 0 + 3 \cdot 2 = 2 + 0 + 6 = 8 \)
- Third row, second column: \( 2 \cdot 0 + 0 \cdot 2 + 3 \cdot 0 = 0 + 0 + 0 = 0 \)
- Third row, third column: \( 2 \cdot 2 + 0 \cdot 1 + 3 \cdot 3 = 4 + 0 + 9 = 13 \)
Thus,
\[
A^2 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix}
\]
### Step 2: Calculate \( A^3 \)
Now we calculate \( A^3 \) by multiplying \( A \) with \( A^2 \):
\[
A^3 = A \cdot A^2 = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix}
\]
Calculating the entries of \( A^3 \):
- First row, first column: \( 1 \cdot 5 + 0 \cdot 2 + 2 \cdot 8 = 5 + 0 + 16 = 21 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 4 + 2 \cdot 0 = 0 + 0 + 0 = 0 \)
- First row, third column: \( 1 \cdot 8 + 0 \cdot 5 + 2 \cdot 13 = 8 + 0 + 26 = 34 \)
- Second row, first column: \( 0 \cdot 5 + 2 \cdot 2 + 1 \cdot 8 = 0 + 4 + 8 = 12 \)
- Second row, second column: \( 0 \cdot 0 + 2 \cdot 4 + 1 \cdot 0 = 0 + 8 + 0 = 8 \)
- Second row, third column: \( 0 \cdot 8 + 2 \cdot 5 + 1 \cdot 13 = 0 + 10 + 13 = 23 \)
- Third row, first column: \( 2 \cdot 5 + 0 \cdot 2 + 3 \cdot 8 = 10 + 0 + 24 = 34 \)
- Third row, second column: \( 2 \cdot 0 + 0 \cdot 4 + 3 \cdot 0 = 0 + 0 + 0 = 0 \)
- Third row, third column: \( 2 \cdot 8 + 0 \cdot 5 + 3 \cdot 13 = 16 + 0 + 39 = 55 \)
Thus,
\[
A^3 = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix}
\]
### Step 3: Substitute into the equation
Now we substitute \( A^3 \), \( A^2 \), and \( A \) into the equation:
\[
A^3 - 6A^2 + 7A + kI_3 = O
\]
Calculating \( -6A^2 \):
\[
-6A^2 = -6 \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} = \begin{pmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{pmatrix}
\]
Calculating \( 7A \):
\[
7A = 7 \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix}
\]
Calculating \( kI_3 \):
\[
kI_3 = k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix}
\]
### Step 4: Combine all parts
Now we combine all parts:
\[
\begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} + \begin{pmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{pmatrix} + \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix} + \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
\]
Calculating each entry:
1. First row:
- First column: \( 21 - 30 + 7 + k = -2 + k \)
- Second column: \( 0 + 0 + 0 = 0 \)
- Third column: \( 34 - 48 + 14 = 0 \)
2. Second row:
- First column: \( 12 - 12 + 0 + 0 = 0 \)
- Second column: \( 8 - 24 + 14 + k = -2 + k \)
- Third column: \( 23 - 30 + 7 = 0 \)
3. Third row:
- First column: \( 34 - 48 + 14 = 0 \)
- Second column: \( 0 + 0 + 0 = 0 \)
- Third column: \( 55 - 78 + 21 + k = -2 + k \)
### Step 5: Set equations to zero
From the first row, first column:
\[
-2 + k = 0 \implies k = 2
\]
From the second row, second column:
\[
-2 + k = 0 \implies k = 2
\]
From the third row, third column:
\[
-2 + k = 0 \implies k = 2
\]
### Conclusion
Thus, the value of \( k \) is:
\[
\boxed{2}
\]
