Solution
= Solution
The <determinant> is $2\cdot4-3\cdot1=5\ne0$, so the <matrix inverse> exists. Using the two-by-two <adjugate matrix> formula gives
$$
\boxed{A^{-1}=\frac15\begin{pmatrix}4&-1\\-3&2\end{pmatrix}}.
$$
Multiplication on either side gives the <identity matrix>, which also checks the signs of the off-diagonal entries.