Let `A=[(1,1,3),(5,2,6),(-2,-1,-3)]` then A is

Options

A. Nilpotent
B. Idempotent
C. Scalar
D. None of these

Answer

To determine the nature of the matrix \( A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \), we need to find \( A^2 \) and \( A^3 \) and check if \( A^2 = 0 \) or \( A^3 = 0 \).

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \times A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \times \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \]

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

- First row, first column:
\[ 1 \cdot 1 + 1 \cdot 5 + 3 \cdot (-2) = 1 + 5 - 6 = 0 \]

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

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

- Second row, first column:
\[ 5 \cdot 1 + 2 \cdot 5 + 6 \cdot (-2) = 5 + 10 - 12 = 3 \]

- Second row, second column:
\[ 5 \cdot 1 + 2 \cdot 2 + 6 \cdot (-1) = 5 + 4 - 6 = 3 \]

- Second row, third column:
\[ 5 \cdot 3 + 2 \cdot 6 + 6 \cdot (-3) = 15 + 12 - 18 = 9 \]

- Third row, first column:
\[ -2 \cdot 1 + (-1) \cdot 5 + (-3) \cdot (-2) = -2 - 5 + 6 = -1 \]

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

- 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 \( A^3 = A^2 \times A \):

\[ A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{pmatrix} \times \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \]

Calculating each element of \( A^3 \):

- First row, first column:
\[ 0 \cdot 1 + 0 \cdot 5 + 0 \cdot (-2) = 0 \]

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

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

- Second row, first column:
\[ 3 \cdot 1 + 3 \cdot 5 + 9 \cdot (-2) = 3 + 15 - 18 = 0 \]

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

- Second row, third column:
\[ 3 \cdot 3 + 3 \cdot 6 + 9 \cdot (-3) = 9 + 18 - 27 = 0 \]

- Third row, first column:
\[ -1 \cdot 1 + (-1) \cdot 5 + (-3) \cdot (-2) = -1 - 5 + 6 = 0 \]

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

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

Thus, we have:

\[ A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Conclusion

Since \( A^3 = 0 \), matrix \( A \) is nilpotent.

### Final Answer

**A is nilpotent.**