The matrix `{:A=[(1,-3,-4),(-1,3,4),(1,-3,-4)]:}`is nilpotent of index

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To determine the index of the nilpotent matrix \( A \), we need to find the smallest positive integer \( n \) such that \( A^n = 0 \) (the null matrix).

Given the matrix:
\[ A = \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \]

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

To find \( A^2 \), we perform matrix multiplication of \( A \) with itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \cdot \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \]

Calculating each element of the resulting matrix:

1. **First row, first column**:
\[ 1 \cdot 1 + (-3) \cdot (-1) + (-4) \cdot 1 = 1 + 3 - 4 = 0 \]

2. **First row, second column**:
\[ 1 \cdot (-3) + (-3) \cdot 3 + (-4) \cdot (-3) = -3 - 9 + 12 = 0 \]

3. **First row, third column**:
\[ 1 \cdot (-4) + (-3) \cdot 4 + (-4) \cdot (-4) = -4 - 12 + 16 = 0 \]

4. **Second row, first column**:
\[ (-1) \cdot 1 + 3 \cdot (-1) + 4 \cdot 1 = -1 - 3 + 4 = 0 \]

5. **Second row, second column**:
\[ (-1) \cdot (-3) + 3 \cdot 3 + 4 \cdot (-3) = 3 + 9 - 12 = 0 \]

6. **Second row, third column**:
\[ (-1) \cdot (-4) + 3 \cdot 4 + 4 \cdot (-4) = 4 + 12 - 16 = 0 \]

7. **Third row, first column**:
\[ 1 \cdot 1 + (-3) \cdot (-1) + (-4) \cdot 1 = 1 + 3 - 4 = 0 \]

8. **Third row, second column**:
\[ 1 \cdot (-3) + (-3) \cdot 3 + (-4) \cdot (-3) = -3 - 9 + 12 = 0 \]

9. **Third row, third column**:
\[ 1 \cdot (-4) + (-3) \cdot 4 + (-4) \cdot (-4) = -4 - 12 + 16 = 0 \]

Thus, we find that:
\[ A^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Conclusion

Since \( A^2 = 0 \), the matrix \( A \) is nilpotent of index 2.