Given,

The matrix $\mathrm{A}=\left[\begin{array}{cc}2& -3\\ 4& 6\end{array}\right]$

The determinant of A, |A| = (2 × 6) – (– 3 × 4) = 12 + 12 = 24 ≠ 0

This indicates that the matrix have inverse

In order to find adjoint of the matrix A let's find the minors of the matrix A $=\left[\begin{array}{cc}\left|6\right|& \left|4\right|\\ |-3|& \left|2\right|\end{array}\right]$

Co-factors

$=\left[\begin{array}{cc}(-1{)}^{1+1}(6)& (-1{)}^{1+2}(4)\\ (-1{)}^{2+1}(-3)& (-1{)}^{2+2}(2)\end{array}\right]=\left[\begin{array}{cc}6& -4\\ 3& 2\end{array}\right]$

Transpose of the matrix A with co-factors

${\mathrm{C}}_{\mathrm{A}}^{\mathrm{T}}={\left[\begin{array}{cc}6& -4\\ 3& 2\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{cc}6& 3\\ -4& 2\end{array}\right]=\mathrm{adj}\mathrm{A}$

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

$=\frac{1}{24}\left[\begin{array}{cc}6& 3\\ -4& 2\end{array}\right]$

${\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{8}\\ \frac{-1}{6}& \frac{1}{12}\end{array}\right]$