# How do you find the inverse of A=((1, 1, -1), (0, 1, 0), (1, 0, 1))?

Dec 16, 2016

$\implies {A}^{- 1} = \left(\begin{matrix}\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ - \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{matrix}\right)$

#### Explanation:

Write as:

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

$\left(\begin{matrix}0 & 0 & - 2 & \text{|" & 1 & -1 & -1 \\ 0 & 1 & 0 & "|" & 0 & 1 & 0 \\ 1 & 0 & 1 & "|} & 0 & 0 & 1\end{matrix}\right)$
$\text{ } {R}_{1} \div \left(- 2\right)$
$\text{ } \downarrow$

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

$\left(\begin{matrix}0 & 0 & 1 & \text{|" & -1/2 & 1/2 & 1/2 \\ 0 & 1 & 0 & "|" & 0 & 1 & 0 \\ 1 & 0 & 0 & "|} & \frac{1}{2} & - \frac{1}{2} & \frac{1}{2}\end{matrix}\right)$
$\text{ Swap "R_3" with } {R}_{1}$
$\text{ } \downarrow$

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

$\textcolor{w h i t e}{.}$

$\implies {A}^{- 1} = \left(\begin{matrix}\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ - \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{matrix}\right)$