<p>Let \(A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\). The only correct statement about the matrix \(A\) is</p>
<p>\(A\) is a zero matrix</p>
<p>\(A^2 = I\)</p>
<p>\(A^{-1}\) does not exist</p>
<p>\(A = (-1)\,I\), where \(I\) is a unit matrix</p>
Step-by-Step Solution
Key Concept: Recognize that A is an involutory matrix (A² = I) by computing A·A, and this special property constrains all possible statements about eigenvalues, determinant, and powers of A.
<p><strong>Step 1:</strong> Compute $A^2$</p><p>$A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I$</p><p><strong>Step 2:</strong> Deduce properties from $A^2 = I$</p><ul><li>$A$ is involutory (applying to itself gives the identity)</li><li>$\det(A^2) = \det(I) \implies (\det A)^2 = 1 \implies \det(A) = \pm 1$. Computing: $\det(A) = 0 \cdot (0) - 0 \cdot (0) + (-1) \cdot (0-1) = -1$</li><li>From $A^2 = I$: eigenvalues $\lambda$ satisfy $\lambda^2 = 1$, so $\lambda \in \{1, -1\}$</li><li>$A^n$ alternates: $A^{\text{odd}} = A$, $A^{\text{even}} = I$</li><li>$A = A^{-1}$ (since $AA = I$)</li></ul><p><strong>Step 3:</strong> Identify the correct statement among options</p><p>The only correct statement must reflect one of: $\det(A) = -1$, $A^2 = I$, eigenvalues are $\pm 1$, or $A$ is involutory.</p><p>$\therefore$ Answer: <strong>B</strong> ($A^2 = I$)</p>
Correct Answer: B