Step 1: Set up:
The matrix $A$ has entries $-1$ at positions $(1,3),(2,2),(3,1)$ and zeros elsewhere. Each row has a single nonzero entry, so products are easy.
Step 2: Treat it as a signed permutation:
Write $A = -P$, where $P$ swaps the first and third coordinates and keeps the second. A swap applied twice gives back the identity, so $P^2 = I$. Then $A^2 = (-P)^2 = P^2 = I$.
Step 3: Inverse:
From $A^2 = I$ we get $A^{-1} = A$. Option (C) is correct.
Step 4: Test the distractors with the determinant:
$|A| = |-P| = (-1)^3 |P| = (-1)^3(-1) = 1$, because $P$ is one swap and has determinant $-1$. So $\text{adj}A = |A|A^{-1} = A$. Option (B) says $A = -\text{adj}A$, which would force $A = -A$, so it is false. Options (A) and (D) fail since $A^2 = I \neq A$ and $-A \neq A$.
Final Answer:
Since $A^2=I$, $A^{-1}=A$.
\[ \boxed{A^{-1}=A} \]