# How do you solve using gaussian elimination or gauss-jordan elimination, #2x-y-z=9#, #3x+2y+z=17#, #x+2y+2z=7#?

##### 1 Answer

First, let's put this into an **augmented matrix**, where the last column contains the answers to the system represented by the left columns.

#[(2,-1,-1,|,9),(3,2,1,|,17),(1,2,2,|,7)]#

Now, let's go for **Gauss-Jordan elimination**, since that covers both row echelon and reduced-row echelon forms.

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

We can use **elementary row operations** to achieve this. Common ones are:

- Scaling a row
- Swapping two rows
- Adding/subtracting two rows, even if one row is scaled as well

I will use the notation where the rightmost indicated row is where the operation occurs.

#stackrel(-R_3 + R_1" ")(->)[(1,-3,-3,|,2),(3,2,1,|,17),(1,2,2,|,7)]#

#stackrel(-3R_3 + R_2" ")(->)[(1,-3,-3,|,2),(0,-4,-5,|,-4),(1,2,2,|,7)]#

#stackrel(-R_1 + R_3" ")(->)[(1,-3,-3,|,2),(0,-4,-5,|,-4),(0,5,5,|,5)]#

#stackrel(R_3 + R_2" ")(->)[(1,-3,-3,|,2),(0,1,0,|,1),(0,5,5,|,5)]#

#stackrel(1/5R_3" ")(->)[(1,-3,-3,|,2),(0,1,0,|,1),(0,1,1,|,1)]#

#stackrel(-R_2 + R_3" ")(->)[(1,-3,-3,|,2),(0,1,0,|,1),(0,0,1,|,0)]#

#stackrel(3R_3 + R_1; 3R_2 + R_1" ")(->)color(green)([(1,0,0,|,5),(0,1,0,|,1),(0,0,1,|,0)])#

At this point, we can reconstruct the system of equations to get:

#color(blue)((x,y,z) = (5,1,0))#

If we check with the original equations, we get:

#2(5) - 1(1) - 1(0) = 9#

#3(5) + 2(1) + 1(0) = 17#

#1(5) + 2(1) + 2(0) = 7#