To determine the correct classification of the matrix [tex]\( A \)[/tex], we need to analyze its properties. We are given the matrix
[tex]\[ A = \left[\begin{array}{ccc}
0 & 0 & 1 \\
0 & -1 & 0 \\
1 & 0 & 0
\end{array}\right]. \][/tex]
We need to check if this matrix [tex]\( A \)[/tex] fits any of the given options:
1. Diagonal matrix:
- A diagonal matrix is one where all the non-diagonal elements are zero.
- For matrix [tex]\( A \)[/tex], the non-diagonal elements are not all zero. Hence, [tex]\( A \)[/tex] is not a diagonal matrix.
2. Idempotent matrix:
- A matrix [tex]\( B \)[/tex] is idempotent if [tex]\( B^2 = B \)[/tex].
- We need to check if [tex]\( A^2 = A \)[/tex]. Without performing the matrix multiplication, one observation is enough to show that [tex]\( A \neq A^2 \)[/tex] just by checking the structure. Therefore [tex]\( A \)[/tex] is not an idempotent matrix.
3. Involutory matrix:
- A matrix [tex]\( C \)[/tex] is involutory if [tex]\( C^2 = I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix.
- To determine if [tex]\( A \)[/tex] is involutory, we calculate [tex]\( A^2 \)[/tex] and check if it equals the identity matrix.
4. Singular matrix:
- A matrix is singular if its determinant is zero.
- We calculate the determinant of [tex]\( A \)[/tex]:
[tex]\[
\text{det}(A) = \begin{vmatrix}
0 & 0 & 1 \\
0 & -1 & 0 \\
1 & 0 & 0
\end{vmatrix} = 0
\][/tex]
- From the determinant calculation, [tex]\(\text{det}(A) \neq 0\)[/tex]. Thus, [tex]\( A \)[/tex] is not a singular matrix.
Given the checks and properties:
We observe that to determine if [tex]\( A \)[/tex] is involutory, we verified that [tex]\( A^2 \)[/tex] does indeed equal the identity matrix [tex]\( I \)[/tex]. Therefore, the correct answer is:
[tex]\[ A \text{ is an }\textbf{involutory matrix} \][/tex]
Thus, the correct answer is:
C. Involutory matrix
