Question

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

Answer 1

The solution is `((x),(y),(y))=((2/3),(1),(-2/3))`

Explanation:

Perform the Jordan-Gauss elimination on the augmented matrix

`A=((1,1,1,|,1),(1,1,-2,|,3),(1,2,1,|,2))`

The pivot is in the first colum and the first row.

Eliminate the first column by performing the row operations

`R2larrR2-R1` and `R3larrR3-R1`

`((1,1,1,|,1),(0,0,-3,|,2),(0,1,0,|,1))`

Swap the last two rows

`R2harrR3`

`((1,1,1,|,1),(0,1,0,|,1),(0,0,-3,|,2))`

The pivot is in the second row and second column

Eliminate the second column

`R1larrR1-R2`

`((1,0,1,|,0),(0,1,0,|,1),(0,0,-3,|,2))`

Make the pivot in the third row of third column

`R3larrR3*(-1/3)`

`((1,0,1,|,0),(0,1,0,|,1),(0,0,1,|,-2/3))`

Eliminate the third column

`R1larrR1-R3`

`((1,0,0,|,2/3),(0,1,0,|,1),(0,0,1,|,-2/3))`

The solution is

`((x),(y),(y))=((2/3),(1),(-2/3))`