Some special square matrices are defined as follows. Nilpotent matrix: A square matrix. A is said to be nilpotent (of order 2)if, `A^(2)=O`. A square matrix is said to be nilpotent of order p, if p is the least positive integer such that `A^(p)=O`.
Idempotent matrix: A square matrix A is said to be idempotent if, `A^(2)=A`.
`e.g.[{:(,1,0),(,0,1):}]` is an idempotent matrix.
Involutory matrix: A square A is said to be involutary if `A^(2)=I`, `I `being the identity matrix.
`e.g..A=[{:(,1,0),(,0,1):}]` is an involutary matrix.
Orthogonal matrix: A square matrix A is said to be an orthogonal matrix if `A' A=I=A A'`
The matrix A `[{:(,1,1,3),(,5,2,6),(,-2,-1,-3):}]`

Options

A. idempotent matrices
B. involutary matrices
C. nilpotent matrix
D. none of these

Answer

To determine the type of the given matrix \( A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \), we will follow these steps:

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

We need to multiply matrix \( A \) by itself to find \( A^2 \).

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

Calculating the elements of \( A^2 \):

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:
\[ -2 \cdot 3 + -1 \cdot 6 + -3 \cdot (-3) = -6 - 6 + 9 = -3 \]

So, we have:

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

### Step 2: Check if \( A^2 = O \)

We need to check if \( A^2 \) equals the zero matrix \( O \).

Since \( A^2 \) is not the zero matrix, we will calculate \( A^3 \).

### Step 3: Calculate \( A^3 \)

Now, we will calculate \( A^3 = A^2 \cdot A \):

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

Calculating the elements of \( A^3 \):

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

So, the first row will be \( (0, 0, 0) \).

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

So, we have:

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

### Conclusion

Since \( A^3 = O \) and \( A^2 \neq O \), matrix \( A \) is nilpotent of order 3.

### Final Answer

The matrix \( A \) is a nilpotent matrix of order 3.

---