Given matrix $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$
Then determinant of the matrix, $\mathrm{A}=\left|\mathrm{A}\right|=\left|\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right|=\text{ ad-bc }\ne 0$
This indicates that there exists inverse of the matrix A.
In order to find the adjoint let's find minor matrix which is given as,

$=\left(\begin{array}{l}\left|\mathrm{d}\right|\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|\mathrm{c}\right|\\ \left|\mathrm{b}\right|\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|\mathrm{a}\right|\end{array}\right)=\left[\begin{array}{l}\mathrm{d}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}\\ \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}\end{array}\right]$

Co-factors of matrix $\mathrm{A}=\left[\begin{array}{l}(-1{)}^{1+1}(\mathrm{d})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-1{)}^{1+2}(\mathrm{c})\\ (-1{)}^{2+1}(\mathrm{b})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-1{)}^{2+2}(\mathrm{a})\end{array}\right]=\left[\begin{array}{cc}\mathrm{d}& -\mathrm{c}\\ -\mathrm{b}& \mathrm{a}\end{array}\right]$

Transpose of matrix A with co-factors, $={{\mathrm{C}}_{\mathrm{A}}}^{\mathrm{T}}={\left[\begin{array}{cc}\mathrm{d}& -\mathrm{c}\\ -\mathrm{b}& \mathrm{a}\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]$

Then inverse of the matrix $\mathrm{A},{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{adj}\mathrm{A}$

$=\frac{1}{\mathrm{ad}-\mathrm{bc}}\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]$