# How do you solve using gaussian elimination or gauss-jordan elimination, #-x+y-z=1#, #-x+3y+z=3#, #x+2y+4z=2#?

##### 1 Answer

First, let's convert this to an **augmented matrix** form because I find it faster to write out.

#[(-1,1,-1,|,1),(-1,3,1,|,3),(1,2,4,|,2)]#

Each number is a coefficient in the original system of equations, except for the last column, which contains the answers to the equalities.

I've done an answer already where I showed Gauss-Jordan elimination, but this time let's do **Gaussian elimination**, which we just perform elementary row operations until we get **row echelon form** instead of reduced-row echelon form. Essentially, it's fewer steps, but is still solvable (if and only if there is **one** solution).

The goal for **row echelon form** is to get the first nonzero entry in each row to be a

(For **reduced-row echelon form**, go further and achieve

*The notation I will use is that the rightmost row in the step is the one operated upon.*

Here we go!

#stackrel(-R_1" ")(->)[(1,-1,1,|,-1),(-1,3,1,|,3),(1,2,4,|,2)]#

#stackrel(R_3 + R_2" ")(->)[(1,-1,1,|,-1),(0,5,5,|,5),(1,2,4,|,2)]#

#stackrel(1/5R_2" ")(->)[(1,-1,1,|,-1),(0,1,1,|,1),(1,2,4,|,2)]#

#stackrel(-R_1 + R_3" ")(->)[(1,-1,1,|,-1),(0,1,1,|,1),(0,3,3,|,3)]#

Oh boy. Yeah, this has infinite solutions. Why? Because at this point,

#stackrel(-3R_2 + R_3" ")(->)[(1,-1,1,|,-1),(0,1,1,|,1),(0,0,0,|,0)]#

We lost an equation, so we have **three** variables and **two** equations. Not good. When you turn this back to algebraic form, you get:

#x - y + z = -1#

#y + z = 1#

If I attempt to solve this, I get:

#y = 1 - z#

#x - (1 - z) + z = -1#

#x - 1 + 2z = -1#

#color(red)(z = -x/2)#

#color(red)(y = 1 + x/2)#

But **this system has infinite solutions.**