Home > Matrix & Vector calculators > Inverse of matrix using Gauss-Jordan Elimination method example

2. Gauss-Jordan Elimination method example ( 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. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
(Previous example)
3. Example `[[2,3],[4,10]]`
(Next example)

2. Example `[[2,3,1],[0,5,6],[1,1,2]]`





2. Find Inverse of matrix using Gauss-Jordan Elimination method
`A=[[2,3,1],[0,5,6],[1,1,2]]`


Solution:
Given matrix is
`2``3``1`
`0``5``6`
`1``1``2`


Now finding inverse of the given matrix
`2``3``1``1``0``0`
`0``5``6``0``1``0`
`1``1``2``0``0``1`


`R_1 larr R_1-:2`

 = 
 `1` `1=2-:2`
`R_1 larr R_1-:2`
 `3/2` `3/2=3-:2`
`R_1 larr R_1-:2`
 `1/2` `1/2=1-:2`
`R_1 larr R_1-:2`
 `1/2` `1/2=1-:2`
`R_1 larr R_1-:2`
 `0` `0=0-:2`
`R_1 larr R_1-:2`
 `0` `0=0-:2`
`R_1 larr R_1-:2`
`0``5``6``0``1``0`
`1``1``2``0``0``1`


`R_3 larr R_3- R_1`

 = 
`1``3/2``1/2``1/2``0``0`
`0``5``6``0``1``0`
 `0` `0=1-1`
`R_3 larr R_3- R_1`
 `-1/2` `-1/2=1-3/2`
`R_3 larr R_3- R_1`
 `3/2` `3/2=2-1/2`
`R_3 larr R_3- R_1`
 `-1/2` `-1/2=0-1/2`
`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_2-:5`

 = 
`1``3/2``1/2``1/2``0``0`
 `0` `0=0-:5`
`R_2 larr R_2-:5`
 `1` `1=5-:5`
`R_2 larr R_2-:5`
 `6/5` `6/5=6-:5`
`R_2 larr R_2-:5`
 `0` `0=0-:5`
`R_2 larr R_2-:5`
 `1/5` `1/5=1-:5`
`R_2 larr R_2-:5`
 `0` `0=0-:5`
`R_2 larr R_2-:5`
`0``-1/2``3/2``-1/2``0``1`


`R_1 larr R_1-3/2xx R_2`

 = 
 `1` `1=1-3/2xx0`
`R_1 larr R_1-3/2xx R_2`
 `0` `0=3/2-3/2xx1`
`R_1 larr R_1-3/2xx R_2`
 `-13/10` `-13/10=1/2-3/2xx6/5`
`R_1 larr R_1-3/2xx R_2`
 `1/2` `1/2=1/2-3/2xx0`
`R_1 larr R_1-3/2xx R_2`
 `-3/10` `-3/10=0-3/2xx1/5`
`R_1 larr R_1-3/2xx R_2`
 `0` `0=0-3/2xx0`
`R_1 larr R_1-3/2xx R_2`
`0``1``6/5``0``1/5``0`
`0``-1/2``3/2``-1/2``0``1`


`R_3 larr R_3+1/2xx R_2`

 = 
`1``0``-13/10``1/2``-3/10``0`
`0``1``6/5``0``1/5``0`
 `0` `0=0+1/2xx0`
`R_3 larr R_3+1/2xx R_2`
 `0` `0=-1/2+1/2xx1`
`R_3 larr R_3+1/2xx R_2`
 `21/10` `21/10=3/2+1/2xx6/5`
`R_3 larr R_3+1/2xx R_2`
 `-1/2` `-1/2=-1/2+1/2xx0`
`R_3 larr R_3+1/2xx R_2`
 `1/10` `1/10=0+1/2xx1/5`
`R_3 larr R_3+1/2xx R_2`
 `1` `1=1+1/2xx0`
`R_3 larr R_3+1/2xx R_2`


`R_3 larr R_3xx10/21`

 = 
`1``0``-13/10``1/2``-3/10``0`
`0``1``6/5``0``1/5``0`
 `0` `0=0xx10/21`
`R_3 larr R_3xx10/21`
 `0` `0=0xx10/21`
`R_3 larr R_3xx10/21`
 `1` `1=21/10xx10/21`
`R_3 larr R_3xx10/21`
 `-5/21` `-5/21=-1/2xx10/21`
`R_3 larr R_3xx10/21`
 `1/21` `1/21=1/10xx10/21`
`R_3 larr R_3xx10/21`
 `10/21` `10/21=1xx10/21`
`R_3 larr R_3xx10/21`


`R_1 larr R_1+13/10xx R_3`

 = 
 `1` `1=1+13/10xx0`
`R_1 larr R_1+13/10xx R_3`
 `0` `0=0+13/10xx0`
`R_1 larr R_1+13/10xx R_3`
 `0` `0=-13/10+13/10xx1`
`R_1 larr R_1+13/10xx R_3`
 `4/21` `4/21=1/2+13/10xx-5/21`
`R_1 larr R_1+13/10xx R_3`
 `-5/21` `-5/21=-3/10+13/10xx1/21`
`R_1 larr R_1+13/10xx R_3`
 `13/21` `13/21=0+13/10xx10/21`
`R_1 larr R_1+13/10xx R_3`
`0``1``6/5``0``1/5``0`
`0``0``1``-5/21``1/21``10/21`


`R_2 larr R_2-6/5xx R_3`

 = 
`1``0``0``4/21``-5/21``13/21`
 `0` `0=0-6/5xx0`
`R_2 larr R_2-6/5xx R_3`
 `1` `1=1-6/5xx0`
`R_2 larr R_2-6/5xx R_3`
 `0` `0=6/5-6/5xx1`
`R_2 larr R_2-6/5xx R_3`
 `2/7` `2/7=0-6/5xx-5/21`
`R_2 larr R_2-6/5xx R_3`
 `1/7` `1/7=1/5-6/5xx1/21`
`R_2 larr R_2-6/5xx R_3`
 `-4/7` `-4/7=0-6/5xx10/21`
`R_2 larr R_2-6/5xx R_3`
`0``0``1``-5/21``1/21``10/21`


Solution By Gauss-Jordan Elimination method.
`[[4/21,-5/21,13/21],[2/7,1/7,-4/7],[-5/21,1/21,10/21]]`


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



1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
(Previous example)
3. Example `[[2,3],[4,10]]`
(Next example)





Share this solution or page with your friends.


 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.