Step 1: Compute the determinant using a different expansion (along column 1) as a cross-check:
$|A|=1\begin{vmatrix}4&5\\5&6\end{vmatrix}-2\begin{vmatrix}2&3\\5&6\end{vmatrix}+3\begin{vmatrix}2&3\\4&5\end{vmatrix}=1(24-25)-2(12-15)+3(10-12)=-1+6-6=-1$, matching.
Step 2: Verify the inverse by direct multiplication \(A\cdot A^{-1}\overset{?}{=}I\):
Row 1 of $A$ = $(1,2,3)$ against columns of the candidate inverse $\begin{bmatrix}1&-3&2\\-3&3&-1\\2&-1&0\end{bmatrix}$: col1 $\to1(1)+2(-3)+3(2)=1-6+6=1$; col2 $\to1(-3)+2(3)+3(-1)=-3+6-3=0$; col3 $\to1(2)+2(-1)+3(0)=2-2+0=0$. Row 1 gives $(1,0,0)$ ✓.
Step 3: Spot-check row 2 similarly:
$(2,4,5)$ against col1: $2(1)+4(-3)+5(2)=2-12+10=0$; against col2: $2(-3)+4(3)+5(-1)=-6+12-5=1$; against col3: $2(2)+4(-1)+5(0)=4-4+0=0$. Row 2 gives $(0,1,0)$ ✓.
Final Answer:
The multiplication check confirms the inverse.
\[ \boxed{A^{-1}=\begin{bmatrix}1&-3&2\\-3&3&-1\\2&-1&0\end{bmatrix}} \]