Question

How do you solve using gaussian elimination or gauss-jordan elimination, `x-2y+z=1`, `2x-3y+z=5`, `-x-2y+3z=-13`?

Answer 1

`x = 7+t`, `y=3+t`, `z=t`

Explanation:

Begin by creating the augmented matrix, or a matrix with the `x`, `y`, and `z` coefficients on the left side of a vertical bar, and the constant values on the right side:

`[ [1,-2,1,|,1], [2,-3,1,|,5] , [-1,-2,3,|,-13] ]`

We now perform row operations on the 3 rows in order to reduce the left portion of the augmented matrix to the identity form:

`[ [1,0,0], [0,1,0], [0,0,1] ]`

We can do any of the following:

• Swap any two rows
• Multiply a single row by a non-zero constant value
• Add/Subtract a constant multiple of a row to another row

Using the notation `R_n` to refer to row `n`, here's the steps I'd use:

`[ [1,-2,1,|,1], [2,-3,1,|,5] , [-1,-2,3,|,-13] ] {:(),(-2R_1 rightarrow),(+R_1):}`

`[ [1,-2,1,|,1], [0,1,-1,|,3] , [0,-4,4,|,-12] ] {:(),( rightarrow),(+4R_2):}`

`[ [1,-2,1,|,1], [0,1,-1,|,3] , [0,0,0,|,0] ] {:(+2R_2),( rightarrow),():}`

`[ [1,0,-1,|,7], [0,1,-1,|,3] , [0,0,0,|,0] ]`

We can do nothing further [useful]. Notice the bottom line has all zero values. This indicates that this system of equations does not result in a unique `z` value. In fact, we can translate this reduced row matrix back to a system of three unknowns in two equations:

`{(x,,-z,=,7),(,y,-z,=,3):}`

Any value of `z` chosen (out of infinitely many) will result in a unique solution. For instance, if `z=1`, we can find `x=8` and `y=4`.

It's commonly the case for problems like this with infinite solutions that we define everything in terms of a template variable `t` like so:

`x = 7+t`, `y=3+t`, `z=t`