In case you didn't know . . .

Find the inverse of: .$\displaystyle A \;=\;\begin{bmatrix}a&b\\c&d\end{bmatrix}$

[1] Switch the two on the main diagonal $\displaystyle (a\text{ and }d): \;\begin{bmatrix}d & . \\ . & a\end{bmatrix}$

[2] Change the signs of the other two: .$\displaystyle \begin{bmatrix}. & -b \\ -c & . \end{bmatrix}$

[3] Divide everything by the determinant: .$\displaystyle \begin{vmatrix}a&b\\c&d\end{vmatrix} \:=\:ad - bc$

$\displaystyle \text{Therefore: }\:A^{-1} \;=\;\begin{bmatrix}\dfrac{d}{ad-bc} & \dfrac{-b}{ad-bc} \\ \dfrac{-c}{ad-bc} & \dfrac{a}{ad-bc}\end{bmatrix}$