Question

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

Answer 1

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 `1` and entries below each `1` to be `0`, while any zeroed row is at the bottom, and the leading `1` in each successive row is at least one column to the right of the leading `1` in the preceding row.

For reduced-row echelon form, go further and achieve `0`'s above and below all leading `1`'s.

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`