We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more   

AtoZmath.com
Support us
(New) All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home What's new College Algebra Games Feedback About us
Algebra Matrix & Vector Numerical Methods Statistical Methods Operation Research Word Problems Calculus Geometry Pre-Algebra


Home > Matrix & Vector calculators > Inverse of matrix using Gauss-Jordan Elimination method example
2. Gauss-Jordan Elimination method example ( Enter your problem )
( Enter your problem )
  1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
  2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
  3. Example `[[2,3],[4,10]]`
  4. Example `[[5,1],[4,2]]`
 		Other related methods
  1. Adjoint method
  2. Gauss-Jordan Elimination method
  3. Cayley Hamilton method
1. Adjoint method
(Previous method)		2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
(Next example)

1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`


1. Find Inverse of matrix using Gauss-Jordan Elimination method
`A=[[3,1,1],[-1,2,1],[1,1,1]]`

Solution:
Given matrix is
`3``1``1`
`-1``2``1`
`1``1``1`


Now finding inverse of the given matrix
`3``1``1``1``0``0`
`-1``2``1``0``1``0`
`1``1``1``0``0``1`


`R_1 larr R_1-:3`

 = 
 `1` `1=3-:3`
`R_1 larr R_1-:3`		 `1/3` `1/3=1-:3`
`R_1 larr R_1-:3`		 `1/3` `1/3=1-:3`
`R_1 larr R_1-:3`		 `1/3` `1/3=1-:3`
`R_1 larr R_1-:3`		 `0` `0=0-:3`
`R_1 larr R_1-:3`		 `0` `0=0-:3`
`R_1 larr R_1-:3`
`-1``2``1``0``1``0`
`1``1``1``0``0``1`


`R_2 larr R_2+ R_1`

 = 
`1``1/3``1/3``1/3``0``0`
 `0` `0=-1+1`
`R_2 larr R_2+ R_1`		 `7/3` `7/3=2+1/3`
`R_2 larr R_2+ R_1`		 `4/3` `4/3=1+1/3`
`R_2 larr R_2+ R_1`		 `1/3` `1/3=0+1/3`
`R_2 larr R_2+ R_1`		 `1` `1=1+0`
`R_2 larr R_2+ R_1`		 `0` `0=0+0`
`R_2 larr R_2+ R_1`
`1``1``1``0``0``1`


`R_3 larr R_3- R_1`

 = 
`1``1/3``1/3``1/3``0``0`
`0``7/3``4/3``1/3``1``0`
 `0` `0=1-1`
`R_3 larr R_3- R_1`		 `2/3` `2/3=1-1/3`
`R_3 larr R_3- R_1`		 `2/3` `2/3=1-1/3`
`R_3 larr R_3- R_1`		 `-1/3` `-1/3=0-1/3`
`R_3 larr R_3- R_1`		 `0` `0=0-0`
`R_3 larr R_3- R_1`		 `1` `1=1-0`
`R_3 larr R_3- R_1`


`R_2 larr R_2xx3/7`

 = 
`1``1/3``1/3``1/3``0``0`
 `0` `0=0xx3/7`
`R_2 larr R_2xx3/7`		 `1` `1=7/3xx3/7`
`R_2 larr R_2xx3/7`		 `4/7` `4/7=4/3xx3/7`
`R_2 larr R_2xx3/7`		 `1/7` `1/7=1/3xx3/7`
`R_2 larr R_2xx3/7`		 `3/7` `3/7=1xx3/7`
`R_2 larr R_2xx3/7`		 `0` `0=0xx3/7`
`R_2 larr R_2xx3/7`
`0``2/3``2/3``-1/3``0``1`


`R_1 larr R_1-1/3xx R_2`

 = 
 `1` `1=1-1/3xx0`
`R_1 larr R_1-1/3xx R_2`		 `0` `0=1/3-1/3xx1`
`R_1 larr R_1-1/3xx R_2`		 `1/7` `1/7=1/3-1/3xx4/7`
`R_1 larr R_1-1/3xx R_2`		 `2/7` `2/7=1/3-1/3xx1/7`
`R_1 larr R_1-1/3xx R_2`		 `-1/7` `-1/7=0-1/3xx3/7`
`R_1 larr R_1-1/3xx R_2`		 `0` `0=0-1/3xx0`
`R_1 larr R_1-1/3xx R_2`
`0``1``4/7``1/7``3/7``0`
`0``2/3``2/3``-1/3``0``1`


`R_3 larr R_3-2/3xx R_2`

 = 
`1``0``1/7``2/7``-1/7``0`
`0``1``4/7``1/7``3/7``0`
 `0` `0=0-2/3xx0`
`R_3 larr R_3-2/3xx R_2`		 `0` `0=2/3-2/3xx1`
`R_3 larr R_3-2/3xx R_2`		 `2/7` `2/7=2/3-2/3xx4/7`
`R_3 larr R_3-2/3xx R_2`		 `-3/7` `-3/7=-1/3-2/3xx1/7`
`R_3 larr R_3-2/3xx R_2`		 `-2/7` `-2/7=0-2/3xx3/7`
`R_3 larr R_3-2/3xx R_2`		 `1` `1=1-2/3xx0`
`R_3 larr R_3-2/3xx R_2`


`R_3 larr R_3xx7/2`

 = 
`1``0``1/7``2/7``-1/7``0`
`0``1``4/7``1/7``3/7``0`
 `0` `0=0xx7/2`
`R_3 larr R_3xx7/2`		 `0` `0=0xx7/2`
`R_3 larr R_3xx7/2`		 `1` `1=2/7xx7/2`
`R_3 larr R_3xx7/2`		 `-3/2` `-3/2=-3/7xx7/2`
`R_3 larr R_3xx7/2`		 `-1` `-1=-2/7xx7/2`
`R_3 larr R_3xx7/2`		 `7/2` `7/2=1xx7/2`
`R_3 larr R_3xx7/2`


`R_1 larr R_1-1/7xx R_3`

 = 
 `1` `1=1-1/7xx0`
`R_1 larr R_1-1/7xx R_3`		 `0` `0=0-1/7xx0`
`R_1 larr R_1-1/7xx R_3`		 `0` `0=1/7-1/7xx1`
`R_1 larr R_1-1/7xx R_3`		 `1/2` `1/2=2/7-1/7xx-3/2`
`R_1 larr R_1-1/7xx R_3`		 `0` `0=-1/7-1/7xx-1`
`R_1 larr R_1-1/7xx R_3`		 `-1/2` `-1/2=0-1/7xx7/2`
`R_1 larr R_1-1/7xx R_3`
`0``1``4/7``1/7``3/7``0`
`0``0``1``-3/2``-1``7/2`


`R_2 larr R_2-4/7xx R_3`

 = 
`1``0``0``1/2``0``-1/2`
 `0` `0=0-4/7xx0`
`R_2 larr R_2-4/7xx R_3`		 `1` `1=1-4/7xx0`
`R_2 larr R_2-4/7xx R_3`		 `0` `0=4/7-4/7xx1`
`R_2 larr R_2-4/7xx R_3`		 `1` `1=1/7-4/7xx-3/2`
`R_2 larr R_2-4/7xx R_3`		 `1` `1=3/7-4/7xx-1`
`R_2 larr R_2-4/7xx R_3`		 `-2` `-2=0-4/7xx7/2`
`R_2 larr R_2-4/7xx R_3`
`0``0``1``-3/2``-1``7/2`


Solution By Gauss-Jordan Elimination method.
`[[1/2,0,-1/2],[1,1,-2],[-3/2,-1,7/2]]`

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


1. Adjoint method
(Previous method)		2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
(Next example)


Share this solution or page with your friends.


Home What's new College Algebra Games Feedback About us
 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.