A =`[(1,2,3),(1,2,3),(-1,-2,-3)]` then A is a nilpotent matrix of index

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the index of the nilpotent matrix \( A \), we need to compute the powers of \( A \) until we reach the zero matrix. A matrix \( A \) is said to be nilpotent if there exists a positive integer \( k \) such that \( A^k = 0 \), where \( 0 \) is the zero matrix.

Given the matrix:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ -1 & -2 & -3 \end{pmatrix} \]

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

To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ -1 & -2 & -3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ -1 & -2 & -3 \end{pmatrix} \]

Calculating each element of \( A^2 \):

- First row, first column:
\[ 1 \cdot 1 + 2 \cdot 1 + 3 \cdot (-1) = 1 + 2 - 3 = 0 \]

- First row, second column:
\[ 1 \cdot 2 + 2 \cdot 2 + 3 \cdot (-2) = 2 + 4 - 6 = 0 \]

- First row, third column:
\[ 1 \cdot 3 + 2 \cdot 3 + 3 \cdot (-3) = 3 + 6 - 9 = 0 \]

- Second row, first column (same as first row):
\[ 1 \cdot 1 + 2 \cdot 1 + 3 \cdot (-1) = 0 \]

- Second row, second column:
\[ 1 \cdot 2 + 2 \cdot 2 + 3 \cdot (-2) = 0 \]

- Second row, third column:
\[ 1 \cdot 3 + 2 \cdot 3 + 3 \cdot (-3) = 0 \]

- Third row, first column:
\[ -1 \cdot 1 + (-2) \cdot 1 + (-3) \cdot (-1) = -1 - 2 + 3 = 0 \]

- Third row, second column:
\[ -1 \cdot 2 + (-2) \cdot 2 + (-3) \cdot (-2) = -2 - 4 + 6 = 0 \]

- Third row, third column:
\[ -1 \cdot 3 + (-2) \cdot 3 + (-3) \cdot (-3) = -3 - 6 + 9 = 0 \]

Putting it all together, we find:
\[ A^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Conclusion

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

Thus, the answer is:
\[ \text{The index of the nilpotent matrix } A \text{ is } 2. \]