# How do you find the inverse of A=((1, 2), (5, 3))?

Jun 29, 2016

${A}^{- 1} = \left(\begin{matrix}- \frac{3}{7} & \frac{2}{7} \\ \frac{5}{7} & - \frac{1}{7}\end{matrix}\right)$

#### Explanation:

Given any invertible matrix $A$, we can find ${A}^{- 1}$ by creating the augmented matrix $\left(A | I\right)$ and then performing elementary row operations to change the initial matrix into the identity matrix. The result will be $\left(I | {A}^{- 1}\right)$

$\left(A | I\right) = \left(\begin{matrix}1 & 2 & | & 1 & 0 \\ 5 & 3 & | & 0 & 1\end{matrix}\right)$

${R}_{2} - 5 {R}_{1} \to \left(\begin{matrix}1 & 2 & | & 1 & 0 \\ 0 & - 7 & | & - 5 & 1\end{matrix}\right)$

$- \frac{1}{7} {R}_{2} \to \left(\begin{matrix}1 & 2 & | & 1 & 0 \\ 0 & 1 & | & \frac{5}{7} & - \frac{1}{7}\end{matrix}\right)$

${R}_{1} - 2 {R}_{2} \to \left(\begin{matrix}1 & 0 & | & - \frac{3}{7} & \frac{2}{7} \\ 0 & 1 & | & \frac{5}{7} & - \frac{1}{7}\end{matrix}\right) = \left(I | {A}^{- 1}\right)$

Thus we have ${A}^{- 1} = \left(\begin{matrix}- \frac{3}{7} & \frac{2}{7} \\ \frac{5}{7} & - \frac{1}{7}\end{matrix}\right)$

Verifying this, we find that $A {A}^{- 1} = I$