Question

How do you solve using gaussian elimination or gauss-jordan elimination, `3y + 2z = 4`, `2x − y − 3z = 3`, `2x+ 2y − z = 7`?

Answer 1

`{((13+7p)/6, (4-2p)/3, p), p in RR}`

Explanation:

`0x+3y+2z=4`
`2x−y−3z=3`
`2x+2y−z=7`

`((0,3,2quadquad|4),(2,-1,-3|3),(2,2,-1|7))~~((2,-1,-3|3),(2,2,-1|7),(0,3,2quadquad|4))~~`
`R_3hArrR_1`
`R_2=R_2-R_1`

`~~((2,-1,-3|3),(0,3,2quad|4),(0,3,2quad|4))~~((2,-1,-3|3),(0,3,2quad|4),(0,0,0quad|0))~~`
`R_3=R_3-R_2`
`R_2=1/3*R_2`

`~~((2,-1,-3|3),(0,1,2/3quad|4/3),(0,0,0quad|0))~~((2,0,-7/3|13/3),(0,1,2/3quad|4/3),(0,0,0quad|0))~~((1,0,-7/6|13/6),(0,1,2/3quad|4/3),(0,0,0quad|0))`
`R_1=R_1+R_2`
`R_1=1/2*R_1`

`x-7/6z=13/6`

`y+2/3z=4/3`

Last row has all zeros which means we have to substitute for parameter. z seems to be the best way to go.

`z=p` (p as parameter)

`x-7/6p=13/6quad=>quadx=13/6+7/6p=(13+7p)/6`

`y+2/3p=4/3quad=>quady=4/3-2/3p=(4-2p)/3`

`{((13+7p)/6, (4-2p)/3, p), p in RR}`