If `A=[(-5, -8, 0),(3, 5, 0), (1, 2, -1)]`, then A is

Options

A. idempotent
B. nilpotent
C. involutory
D. periodic

Answer

To determine the type of matrix \( A = \begin{pmatrix} -5 & -8 & 0 \\ 3 & 5 & 0 \\ 1 & 2 & -1 \end{pmatrix} \), we will check if it is idempotent, nilpotent, involutory, or periodic by performing the necessary calculations.

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

To check if \( A \) is idempotent, we need to compute \( A^2 \) (i.e., \( A \times A \)).

\[ A^2 = A \times A = \begin{pmatrix} -5 & -8 & 0 \\ 3 & 5 & 0 \\ 1 & 2 & -1 \end{pmatrix} \times \begin{pmatrix} -5 & -8 & 0 \\ 3 & 5 & 0 \\ 1 & 2 & -1 \end{pmatrix} \]

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

- First row, first column:
\[ (-5)(-5) + (-8)(3) + (0)(1) = 25 - 24 + 0 = 1 \]

- First row, second column:
\[ (-5)(-8) + (-8)(5) + (0)(2) = 40 - 40 + 0 = 0 \]

- First row, third column:
\[ (-5)(0) + (-8)(0) + (0)(-1) = 0 + 0 + 0 = 0 \]

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

- Second row, second column:
\[ (3)(-8) + (5)(5) + (0)(2) = -24 + 25 + 0 = 1 \]

- Second row, third column:
\[ (3)(0) + (5)(0) + (0)(-1) = 0 + 0 + 0 = 0 \]

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

- Third row, second column:
\[ (1)(-8) + (2)(5) + (-1)(2) = -8 + 10 - 2 = 0 \]

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

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \]

### Step 2: Determine the type of matrix

Since \( A^2 = I \), we conclude that \( A \) is an involutory matrix.

### Final Answer
Thus, the matrix \( A \) is **involutory**.

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