# Given A=((-1, 2), (3, 4)) and B=((-4, 3), (5, -2)), how do you find B^-1?

Apr 9, 2016

${B}^{- 1} = \left(\begin{matrix}- \frac{2}{23} & \frac{3}{23} \\ \frac{5}{23} & \frac{4}{23}\end{matrix}\right)$

#### Explanation:

If ${B}^{- 1}$ is the Inverse of $B$
then
$\textcolor{w h i t e}{\text{XXX}} B \times {B}^{- 1} = I$ where $I$ is the identity matrix

Suppose${B}^{- 1} = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

Then
$\textcolor{w h i t e}{\text{XXX}} B \times {B}^{- 1}$
$\textcolor{w h i t e}{\text{XXX}} = \left(\begin{matrix}- 4 & 3 \\ 5 & 2\end{matrix}\right) \times \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

$\Rightarrow$
$\textcolor{w h i t e}{\text{XXX")-4a+3c=1color(white)("XXX}}$[1]
$\textcolor{w h i t e}{\text{XXX")-4b+3d=0color(white)("XXX}}$[2]
$\textcolor{w h i t e}{\text{XXXX")5a+2c=0color(white)("XXX}}$[3]
$\textcolor{w h i t e}{\text{XXXX")5b+2d=1color(white)("XXX}}$[4]

[1]$\times 5 -$[3]$\times 4$ gives
$\textcolor{w h i t e}{\text{XXX}} 23 c = 5 \rightarrow c = \frac{5}{23}$
and substituting this back into [1] gives
$\textcolor{w h i t e}{\text{XXX}} a = - \frac{2}{23}$

Similarly we can use [2] and [4] to get
$\textcolor{w h i t e}{\text{XXX}} d = \frac{4}{23}$
$\textcolor{w h i t e}{\text{XXX}} b = \frac{2}{23}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {B}^{- 1} = \left(\begin{matrix}- \frac{2}{23} & \frac{3}{23} \\ \frac{5}{23} & \frac{4}{23}\end{matrix}\right)$