Question

Use elementary row operations to find the inverse of \left[\begin{matrix}-2 & -1 & 3 \\ 4 & -2 & 1 \\ 0 & 4 & -2\end{matrix}\right]

I know the inverse as I've done it via cofactor and adjugate. And wolframalpha has confirmed the result. But no matter how many times I try with the elementary row operations I can't get it right.

Answer 1

\[\qquad\qquad\quad\quad[\textbf A|\textbf I]\]\[\left[\begin{matrix}-2 & -1 & 3 \\ 4 & -2 & 1 \\ 0 & 4 & -2\end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 &0 & 1\end{matrix}\right]\]

Answer 2

\[\left[\begin{matrix}-2 & -1 & 3 \\ 0 & -4 & 7 \\ 0 & 4 & -2\end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0 \\ 0 &0 & 1\end{matrix}\right]\qquad R_2\rightarrow R_2+2R_1\]

Answer 3

\[\left[\begin{matrix}-2 & -1 & 3 \\ 0 & -4 & 7 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_3\rightarrow R_3+R_2\]

Answer 4

\[\left[\begin{matrix}2 & 1 & 2 \\ 0 & -4 & 7 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 2 & 1 & 0 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow -(R_1-R_3)\]

Answer 5

\[\left[\begin{matrix}2 & 1 & 2 \\ 0 & 4 & -2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 0 & 0 & 1 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_2\rightarrow -(R_2-R_3)\]

Answer 6

\[\left[\begin{matrix}2 & 1 & 2 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_2\rightarrow R_2/4\]

Answer 7

\[\left[\begin{matrix}2 & 0& 5/2 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 3/4\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow R_1-R_2\]

Answer 8

\[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow R_1/2\]

Answer 9

\[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & -1/2 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 0 & 0 & 1/4 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_3\rightarrow R_3/5\]

Answer 10

\[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & 0 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 1/5 & 1/10 & 7/20 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_2\rightarrow R_2+R_3/2\]

Answer 11

\[\left[\begin{matrix}1 & 0& 0 \\ 0 & 1 & 0 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}0 & 1/4 & 1 /8\\ 1/5 & 1/10 & 7/20 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_1\rightarrow R_1-5 R_3/4\]\[\qquad\qquad[\textbf I |\textbf A^{-1}]\]

Answer 12

wow that is a lot of work^

Answer 13

\[\color{gray} {\begin{matrix}0 & & 0 \\ & & \\ & \smile & \end{matrix}} \]

Answer 14

Wow thank you @UnkleRhaukus !!
I understand why you make below the diagonal 0, but then you started working on the top triangle. Does the order matter? For example can I make the bottom triangle 0 then work down the diagonal making them 1? Is making the bottom triangle 0 the really only required step to be done first?

Answer 15

you can perform any of the allowed row operations at any time and in any order,

some ways are certainly quicker than others ,

im not sure if the steps i used were the simplest, but they did work

Answer 16

Okay, thank you again.

Answer 17

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi