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

Options

A. 3
B. 2
C. 1
D. None of these

Explanation

To determine the nilpotent index of the matrix \( A = \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \), we need to compute the powers of the matrix until we reach the zero matrix. A matrix is nilpotent if there exists a positive integer \( k \) such that \( A^k = 0 \).

### Step-by-Step Solution:

1. **Calculate \( A^2 \)**:
We need to multiply the matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \cdot \begin{pmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{pmatrix} \]

Performing the multiplication:
- First row, first column:
\[ (1)(1) + (-3)(-1) + (-4)(1) = 1 + 3 - 4 = 0 \]
- First row, second column:
\[ (1)(-3) + (-3)(3) + (-4)(-3) = -3 - 9 + 12 = 0 \]
- First row, third column:
\[ (1)(-4) + (-3)(4) + (-4)(-4) = -4 - 12 + 16 = 0 \]
- Second row, first column:
\[ (-1)(1) + (3)(-1) + (4)(1) = -1 - 3 + 4 = 0 \]
- Second row, second column:
\[ (-1)(-3) + (3)(3) + (4)(-3) = 3 + 9 - 12 = 0 \]
- Second row, third column:
\[ (-1)(-4) + (3)(4) + (4)(-4) = 4 + 12 - 16 = 0 \]
- Third row, first column:
\[ (1)(1) + (-3)(-1) + (-4)(1) = 1 + 3 - 4 = 0 \]
- Third row, second column:
\[ (1)(-3) + (-3)(3) + (-4)(-3) = -3 - 9 + 12 = 0 \]
- Third row, third column:
\[ (1)(-4) + (-3)(4) + (-4)(-4) = -4 - 12 + 16 = 0 \]

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

2. **Determine the nilpotent index**:
Since \( A^2 = 0 \), the matrix \( A \) is nilpotent of index 2. This means that the smallest integer \( k \) such that \( A^k = 0 \) is \( k = 2 \).

### Final Answer:
The matrix \( A \) is nilpotent of index 2.