Question

How do you solve using gaussian elimination or gauss-jordan elimination, `2x-y+z=6`, `x+2y-z=1`, `2x-y-z=0`?

Answer 1

Set up an augmented matrix and row-reduce to find
`{(x = 2),(y=1),(z=3):}`

Explanation:

Using Gaussian Elimination on the given system, we start from

`{(2x-y+z=6),(x+2y-z=1),(2x-y-z=0):}`

and use the system to set up the augmented matrix

`((2,-1,1,|,6),(1,2,-1,|,1),(2,-1,-1,|,0))`

Our goal is a matrix of the form

`((1,0,0,|,x),(0,1,0,|,y),(0,0,1,|,z))`

Note that from here, we may use only the following operations:

• multiply a row by a constant

• add a the product of a row and a constant to another row

• switch the positions of two rows

To obtain a `1` in the first row, first column, we multiply the first row by `1/2` and obtain

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

To obtain `0`s for the remainder of the first column, we add `-1` times the first row to the second row, and `-2` times the first row to the third row.

`((1,-1/2,1/2,|,3),(0,5/2,-3/2,|,-2),(0,0,-2,|,-6))`

We could solve for `z` and use back-substitution at this point, but let's proceed with Gauss-Jordan elimination. The next step is to get a `1` in the second row, second column. To do so, we multiply the second row by `2/5`.

`((1,-1/2,1/2,|,3),(0,1,-3/5,|,-4/5),(0,0,-2,|,-6))`

Again, we want `0`s for every other member of the column. To get this, we add `1/2` times the second row to the first row.

`((1,0,1/5,|,13/5),(0,1,-3/5,|,-4/5),(0,0,-2,|,-6))`

As our next to last step, we want a `1` in the third column, third row. To obtain this, we multiply the third row by `-1/2`.

`((1,0,1/5,|,13/5),(0,1,-3/5,|,-4/5),(0,0,1,|,3))`

Finally, we eliminate the remaining nonzero elements of the third column by adding `-1/5` times the third row to the first row, and `3/5` times the third row to the second row.

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

And so we have

`{(x = 2),(y=1),(z=3):}`