To solve the problem, we need to analyze the matrix \( A \) and find the powers of \( A \) until we reach a null matrix. The matrix is given as:
\[
A = \begin{pmatrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{pmatrix}
\]
We need to find the smallest natural number \( x < 15 \) such that \( A^x = O \), where \( O \) is the null matrix.
### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[
A^2 = A \cdot A = \begin{pmatrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{pmatrix} \cdot \begin{pmatrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{pmatrix}
\]
Calculating the elements:
1. First row, first column:
\[
1 \cdot 1 + 1 \cdot 5 + 3 \cdot (-2) = 1 + 5 - 6 = 0
\]
2. First row, second column:
\[
1 \cdot 1 + 1 \cdot 2 + 3 \cdot (-1) = 1 + 2 - 3 = 0
\]
3. First row, third column:
\[
1 \cdot 3 + 1 \cdot 6 + 3 \cdot (-3) = 3 + 6 - 9 = 0
\]
4. Second row, first column:
\[
5 \cdot 1 + 2 \cdot 5 + 6 \cdot (-2) = 5 + 10 - 12 = 3
\]
5. Second row, second column:
\[
5 \cdot 1 + 2 \cdot 2 + 6 \cdot (-1) = 5 + 4 - 6 = 3
\]
6. Second row, third column:
\[
5 \cdot 3 + 2 \cdot 6 + 6 \cdot (-3) = 15 + 12 - 18 = 9
\]
7. Third row, first column:
\[
-2 \cdot 1 + -1 \cdot 5 + -3 \cdot (-2) = -2 - 5 + 6 = -1
\]
8. Third row, second column:
\[
-2 \cdot 1 + -1 \cdot 2 + -3 \cdot (-1) = -2 - 2 + 3 = -1
\]
9. Third row, third column:
\[
-2 \cdot 3 + -1 \cdot 6 + -3 \cdot (-3) = -6 - 6 + 9 = -3
\]
Thus, we have:
\[
A^2 = \begin{pmatrix}
0 & 0 & 0 \\
3 & 3 & 9 \\
-1 & -1 & -3
\end{pmatrix}
\]
### Step 2: Calculate \( A^3 \)
Next, we calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):
\[
A^3 = A^2 \cdot A = \begin{pmatrix}
0 & 0 & 0 \\
3 & 3 & 9 \\
-1 & -1 & -3
\end{pmatrix} \cdot \begin{pmatrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{pmatrix}
\]
Calculating the elements:
1. First row, all columns will be \( 0 \).
2. Second row, first column:
\[
3 \cdot 1 + 3 \cdot 5 + 9 \cdot (-2) = 3 + 15 - 18 = 0
\]
3. Second row, second column:
\[
3 \cdot 1 + 3 \cdot 2 + 9 \cdot (-1) = 3 + 6 - 9 = 0
\]
4. Second row, third column:
\[
3 \cdot 3 + 3 \cdot 6 + 9 \cdot (-3) = 9 + 18 - 27 = 0
\]
5. Third row, first column:
\[
-1 \cdot 1 + -1 \cdot 5 + -3 \cdot (-2) = -1 - 5 + 6 = 0
\]
6. Third row, second column:
\[
-1 \cdot 1 + -1 \cdot 2 + -3 \cdot (-1) = -1 - 2 + 3 = 0
\]
7. Third row, third column:
\[
-1 \cdot 3 + -1 \cdot 6 + -3 \cdot (-3) = -3 - 6 + 9 = 0
\]
Thus, we have:
\[
A^3 = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} = O
\]
### Step 3: Conclusion
Since \( A^3 = O \), we can conclude that \( A^x = O \) for all \( x \geq 3 \) and \( x < 15 \). Therefore, the valid values of \( x \) are \( 3, 4, 5, \ldots, 14 \).
The total number of valid values is \( 14 - 3 + 1 = 12 \).
### Final Answer
The correct option is that the number of values of \( x \) is 12.
