The given matrix is,
[tex]\:A=\:\begin{pmatrix}3&5\\ \:2&4\end{pmatrix}[/tex]
Therefore,
[tex]\begin{gathered} \mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}: \\ \begin{equation*} \quad\begin{pmatrix}a\: & \:b\: \\ c\: & \:d\:\end{pmatrix}^{-1}=\frac{1}{\det\begin{pmatrix}a\: & \:b\: \\ c\: & \:d\:\end{pmatrix}}\begin{pmatrix}d\: & \:-b\: \\ -c\: & \:a\:\end{pmatrix} \end{equation*} \\ =\frac{1}{\det \begin{pmatrix}3&5\\ 2&4\end{pmatrix}}\begin{pmatrix}4&-5\\ -2&3\end{pmatrix} \end{gathered}[/tex]
Where,
[tex]\begin{gathered} \det\begin{pmatrix}3 & 5 \\ 2 & 4\end{pmatrix}=(3\times4)-(2\times5)=12-10=2 \\ \therefore\det\begin{pmatrix}3 & 5 \\ 2 & 4\end{pmatrix}=2 \end{gathered}[/tex]
Hence,
[tex]=\frac{1}{2}\begin{pmatrix}4 & -5 \\ -2 & 3\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\times4 & \frac{1}{2}\times-5 \\ \:\frac{1}{2}\times-2 & \frac{1}{2}\times4\end{pmatrix}=\begin{pmatrix}2 & -\frac{5}{2} \\ -1 & \frac{3}{2}\end{pmatrix}[/tex]
Therefore, the answer is
[tex]A^{-1}=\begin{pmatrix}2 & -\frac{5}{2} \\ -1 & \frac{3}{2}\end{pmatrix}[/tex]
