Step 1: Understand the goal.
We are given $A^{-1}$ and asked for $A$. Since taking the inverse twice returns the original, $A=\left(A^{-1}\right)^{-1}$. So we simply invert the given matrix.
Step 2: Recall the inverse formula.
For any matrix $M$, $M^{-1}=\frac{1}{|M|}\,\text{Adj}(M)$. So we need the determinant and the adjoint of $A^{-1}$.
Step 3: Find the determinant of $A^{-1}$.
Expanding $\left|\begin{smallmatrix}3&2&6\\1&1&2\\2&5&5\end{smallmatrix}\right|$ along the first row: \[ 3(5-10)-2(5-4)+6(5-2)=3(-5)-2(1)+6(3)=-15-2+18=1. \]
Step 4: Use the nice determinant value.
Since $|A^{-1}|=1$, the formula becomes $A=\text{Adj}(A^{-1})$. No division is needed.
Step 5: Find all the cofactors.
$C_{11}=5-10=-5$, $C_{12}=-(5-4)=-1$, $C_{13}=5-2=3$;
$C_{21}=-(10-30)=20$, $C_{22}=15-12=3$, $C_{23}=-(15-4)=-11$;
$C_{31}=4-6=-2$, $C_{32}=-(6-6)=0$, $C_{33}=3-2=1$.
Step 6: Build the adjoint (transpose of cofactors).
Place the cofactors and transpose so they fall into columns: \[ A=\left[\begin{array}{ccc}-5&20&-2\\-1&3&0\\3&-11&1\end{array}\right]. \]
Step 7: Match the option.
This is exactly option (1).
\[ \boxed{A=\left[\begin{array}{ccc}-5&20&-2\\-1&3&0\\3&-11&1\end{array}\right]} \]