Question

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

Answer 1

`x = 1, y = -1, z=1`

Explanation:

To solve the matrix equation `Ax=b` we start by writing the augmented matrix `(A|b)` :
`((1,-2,-1,2),(2,-1,1,4),(-1,1,-2,-4))`
usually we would put a vertical line to separate the fourth column (the column vector `b`) from the first three (which represent `A`, but couldn't figure out how to do it!

The aim in Gauss elimination is to carry out row operations on the augmented matrix until `A` is converted to reduced row-echelon form. To do this we first subtract twice the first row from the second (denote this by `R_2-2R_1` ) and `R_3+R_1`

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

Divide the second row by 3 `(R_2/3)` :

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

`R_3+R_2`

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

`(R_3/{-2})`

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

This completes the Gauss-elimination process. We can now carry out back-substitution to determine the required solution:

`z = 1`

`y+z=0 implies y = -1`

`x-2y-z = 2 implies x = 2y+z+2=2times (-1)+1+2 = 1`

So, the solution is `x = 1, y = -1, z=1`

To carry out Gauss Jordan elimination, the steps are the same until we reach

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

`R_1+2R_2 , R_3+R_2`

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

`(R_3/{-2})`

`((1,0,1,2),(0,1,1,0),(0,0,1,1))`

`R_1-R_3,R_2-R_3`

`((1,0,0,1),(0,1,0,-1),(0,0,1,1))`

from this we can immediately read off `x = 1, y = -1, z=1`