Which one of the following is correct in respect of the matrix ?
`[{:(0,0,-1),(0,-1,0),(-1,0,0):}]` ?

Options

A. A. `A^(-1)` does not exist
B. B. A=(-1) I
C. C. A is a Zero matrix
D. D. `A^(2)=1`

Answer

To determine the correct statement regarding the matrix \( A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} \), we will analyze the properties of this matrix step by step.

### Step 1: Calculate the Determinant of the Matrix \( A \)

To find out if the inverse of the matrix exists, we need to calculate the determinant of \( A \). The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

\[ A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} \]

Here, \( a = 0, b = 0, c = -1, d = 0, e = -1, f = 0, g = -1, h = 0, i = 0 \).

Calculating the determinant:

\[ \text{det}(A) = 0 \cdot ((-1) \cdot 0 - 0 \cdot 0) - 0 \cdot (0 \cdot 0 - 0 \cdot (-1)) + (-1) \cdot (0 \cdot 0 - (-1) \cdot (-1)) \]

This simplifies to:

\[ \text{det}(A) = 0 - 0 - 1 \cdot (0 - 1) = -1 \]

### Step 2: Check if the Inverse Exists

Since the determinant \( \text{det}(A) = -1 \) (which is not equal to 0), we conclude that the inverse of matrix \( A \) exists.

### Step 3: Check if \( A \) is a Scalar Multiple of the Identity Matrix

The identity matrix \( I \) for a \( 3 \times 3 \) matrix is:

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

If \( A \) were equal to \( -1 \cdot I \), it would be:

\[ -1 \cdot I = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \]

Clearly, \( A \) does not equal \( -1 \cdot I \) since the diagonal elements of \( A \) are not all -1.

### Step 4: Check if \( A \) is a Zero Matrix

A zero matrix would have all elements equal to 0:

\[ 0 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

Since \( A \) contains non-zero elements, it is not a zero matrix.

### Step 5: Check if \( A^2 = I \)

To verify if \( A^2 = I \), we calculate \( A \times A \):

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

Calculating the elements:

1. First row, first column: \( 0 \cdot 0 + 0 \cdot 0 + (-1) \cdot (-1) = 1 \)
2. First row, second column: \( 0 \cdot 0 + 0 \cdot (-1) + (-1) \cdot 0 = 0 \)
3. First row, third column: \( 0 \cdot (-1) + 0 \cdot 0 + (-1) \cdot 0 = 0 \)

Continuing this for all elements, we find:

\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \]

### Conclusion

Thus, the correct statement regarding the matrix \( A \) is that \( A^2 = I \).