Answer
$A^{-1}=\dfrac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix} $
Work Step by Step
The determinant for a $2 \times 2$ matrix $\begin{bmatrix} a & b \\c & d \\ \end{bmatrix}$ can be defined as:
$det= \begin{bmatrix} a & b \\c & d \\ \end{bmatrix}=ad-bc$
and the inverse of the matrix $A $ is given by: $A^{-1} =\begin{vmatrix} \dfrac{d}{ad-bc} &\dfrac{-b}{ad-bc}\\
\dfrac{-c}{ad-bc} & \dfrac{a}{ad-bc} \\ \end{vmatrix}$
This can be re-written as: $A^{-1}=\dfrac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix} ~~~(1)$
We know that $|A|=ad-bc$
Now, equation (1) becomes: $A^{-1}=\dfrac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix} $
