# How do you find the inverse of A=((2, -4), (1, 3))?

Feb 16, 2016

${A}^{- 1} = \left(\begin{matrix}\frac{3}{10} & \frac{2}{5} \\ - \frac{1}{10} & \frac{1}{5}\end{matrix}\right)$

#### Explanation:

Extend $A$ with an identity matrix appended on its right:

$\left(\begin{matrix}2 & - 4 & \text{|" & 1 & 0 \\ 1 & 3 & "|} & 0 & 1\end{matrix}\right)$

Perform standard row operations with the objective of modifying the left side into an identity (sub)matrix.
(There are many ways to do this; the following is just one method).

Exchange Rows 1 and 2
$\left(\begin{matrix}1 & 3 & \text{|" & 0 & 1 \\ 2 & -4 & "|} & 1 & 0\end{matrix}\right)$

Subtract $2 \times$ Row 1 from Row 2
$\left(\begin{matrix}1 & 3 & \text{|" & 0 & 1 \\ 0 & -10 & "|} & 1 & - 2\end{matrix}\right)$

Divide Row 2 by $\left(- 10\right)$
$\left(\begin{matrix}1 & 3 & \text{|" & 0 & 1 \\ 0 & 1 & "|} & - \frac{1}{10} & \frac{1}{5}\end{matrix}\right)$

Subtract $3 \times$ Row 2 from Row 1
$\left(\begin{matrix}1 & 0 & \text{|" & 3/10 & 2/5 \\ 0 & 1 & "|} & - \frac{1}{10} & \frac{1}{5}\end{matrix}\right)$

with the identity matrix on the left, the right side is the required inverse.

Of course, it's a good idea to verify this result:
$\left(\begin{matrix}2 & - 4 \\ 1 & 3\end{matrix}\right) \times \left(\begin{matrix}\frac{3}{10} & \frac{2}{5} \\ - \frac{1}{10} & \frac{1}{5}\end{matrix}\right)$

$= \left(\begin{matrix}\frac{6}{10} + \frac{4}{10} & \textcolor{w h i t e}{\text{XXX}} & \frac{4}{5} - \frac{4}{5} \\ \frac{3}{10} - \frac{3}{10} & \null & \frac{2}{5} + \frac{3}{5}\end{matrix}\right)$

$= \left(\begin{matrix}1 & \textcolor{w h i t e}{\text{XX}} & 0 \\ 0 & \null & 1\end{matrix}\right)$
...and we can go home happy!