Question

How do you solve using gaussian elimination or gauss-jordan elimination, `2x + 2y - 3z = -2`, `3x - 1 - 2z = 1`, `2x + 3y - 5z = -3`?

Answer 1

`P=[(-2, 1, 0)]`

Explanation:

I'm assuming that in the second row is missing y. Solving this matrix:

`([2,2,-3,|,-2],[3,-1,-2,|,1],[2,3,-5,|,-3])`
`R_1=R_1xx3`
`R_2=R_2xx-2`
`R_3=R_3xx-3`

`([6,6,-9,|,-6],[-6,2,4,|,-2],[-6,-9,15,|,9])`
`R_2=R_2+R_1`
`R_3=R_3+R_1`

`([6,6,-9,|,-6],[0,8,-5,|,-8],[0,-3,6,|,3])`
`R_2=R_2xx3`
`R_3=R_3xx8`

`([6,6,-9,|,-6],[0,24,-15,|,-24],[0,-24,48,|,24])`
`R_3=R_3+R_2`
`R_1=R_1xx1/3`

`([2,2,-3,|,-2],[0,24,-15,|,-24],[0,0,33,|,0])`
`R_2=R_2xx1/3`
`R_3=R_3xx5/33`

`([2,2,-3,|,-2],[0,8,-5,|,8],[0,0,5,|,0])`
`R_2=R_2+R_3`
`R_3=R_3xx1/5`

`([2,2,-3,|,-2],[0,8,0,|,8],[0,0,1,|,0])`
`R_2=R_2xx1/8`
`R_1=R_1-2xxR_2`
`R_1=R_1+3xxR_3`

`([2,0,0,|,-4],[0,1,0,|,1],[0,0,1,|,0])`
`R_1=R_1xx1/2`

`([1,0,0,|,-2],[0,1,0,|,1],[0,0,1,|,0])`

`P=[(-2, 1, 0)]`

(too long)