# How do you find the inverse of A=((1, 0, -2), (3, 1, -6), (0, 1, 1)) ?

$\left(\begin{matrix}7 & - 2 & 2 \\ - 3 & 1 & 0 \\ 3 & - 1 & 1\end{matrix}\right)$

#### Explanation:

$\left(\begin{matrix}1 & 0 & - 2 \\ 3 & 1 & - 6 \\ 0 & 1 & 1\end{matrix}\right) \cdot \left(\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} \\ {y}_{1} & {y}_{2} & {y}_{3} \\ {z}_{1} & {z}_{2} & {z}_{3}\end{matrix}\right) = \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

${L}_{2} : = {L}_{2} - 3 {L}_{1}$

$\left(\begin{matrix}1 & 0 & - 2 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{matrix}\right) \cdot \left(\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} \\ {y}_{1} & {y}_{2} & {y}_{3} \\ {z}_{1} & {z}_{2} & {z}_{3}\end{matrix}\right) = \left(\begin{matrix}1 & 0 & 0 \\ - 3 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

${L}_{2} : {y}_{1} = - 3 , {y}_{2} = 1 , {y}_{3} = 0$

L_3: y_1 + z_1 = 0 ; y_2 + z_2 = 0; y_3 + z_3 = 1

z_1 = 3 ; z_2 = -1; z_3 = 1

L_1: x_1 - 2 z_1 = 1 ; x_2 - 2 z_2 = 0 ; x_3 - 2 z_3 = 0

x_1 = 7 ; x_2 = - 2 ; x_3 = 2