Answer:
[tex]\therefore A^{-1}=\begin{bmatrix}0.29 & -0.21\\ -0.14 & 0.36\end{bmatrix}[/tex]
Step-by-step explanation:
[tex]A=\begin{bmatrix}5 & 3\\ 2 & 4\end{bmatrix}[/tex]
[tex]det\ A=5\times 4-3\times 2\\\Rightarrow det\ A=14[/tex]
[tex]A^{-1}=\frac{1}{det\ A}\begin{bmatrix}4 & -3\\ -2 & 5\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{14}\begin{bmatrix}4 & -3\\ -2 & 5\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}\frac{2}{7} & \frac{-3}{14}\\ -\frac{1}{7} & \frac{5}{14}\end{bmatrix}[/tex]
[tex]\therefore A^{-1}=\begin{bmatrix}0.29 & -0.21\\ -0.14 & 0.36\end{bmatrix}[/tex]
[tex]A.A^{-1}=\begin{bmatrix}5\:&\:3\\ \:\:2\:&\:4\end{bmatrix}\times \begin{bmatrix}\frac{2}{7}\:&\:\frac{-3}{14}\\ \:\:-\frac{1}{7}\:&\:\frac{5}{14}\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}5\cdot \frac{2}{7}+3\left(-\frac{1}{7}\right)&5\cdot \frac{-3}{14}+3\cdot \frac{5}{14}\\ 2\cdot \frac{2}{7}+4\left(-\frac{1}{7}\right)&2\cdot \frac{-3}{14}+4\cdot \frac{5}{14}\end{pmatrix}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}[/tex]
Hence, proved
