Given ,

The matrix $\mathrm{A}=\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]$

The determinant of A, |A| = (4 × 3) – (1 × 2) = 12 – 2 = 10 ≠ 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

$=\left[\begin{array}{l}\left|3\right|\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|2\right|\\ \left|1\right|\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|4\right|\end{array}\right]$

Co-factors

$=\left[\begin{array}{l}(-1{)}^{1+1}(3)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-1{)}^{1+2}(2)\\ (-1{)}^{2+1}(1)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-1{)}^{2+2}(4)\end{array}\right]=\left[\begin{array}{cc}3& -2\\ -1& 4\end{array}\right]$

Transpose of the matrix A with co-factors

${{\mathrm{C}}_{\mathrm{A}}}^{\mathrm{T}}={\left[\begin{array}{cc}3& -2\\ -1& 4\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{cc}3& -1\\ -2& 4\end{array}\right]=\mathrm{adj}\mathrm{A}$

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

$=\frac{1}{10}\left[\begin{array}{cc}3& -1\\ -2& 4\end{array}\right]$

${\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{3}{10}& \frac{-1}{10}\\ \frac{-2}{10}& \frac{4}{10}\end{array}\right]$