Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$ as $tI - A$ where $t=a+d$ is the trace of $A$. Since $A$ satisfies its own characteristic equation (Cayley-Hamilton), we have $A^2-tA+\Delta A^2 - t A + \Delta \cdot I = 0$ where $\Delta = ad-bc$ is the determinant. Thus $\Delta \cdot I = t A - A^2$. Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$, QED.
Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$ as $tI - A$ where $t=a+d$ is the trace of $A$. Since $A$ satisfies its own characteristic equation, we have $A^2-tA+\Delta \cdot I = 0$ where $\Delta = ad-bc$ is the determinant. Thus $\Delta I = t A - A^2$. Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$, QED.