# Simple Gauss-Jordan Elimination

1. Oct 1, 2009

I have done the following problem over and over and keep getting the same result. So either the book answer is wrong or I am making the same mistake over again:

$$\left[\begin{array} {cccc} 5&-2&6&0\\ -2&1&3&1\\ \end{array}\right]$$

$$\left[\begin{array} {cccc} 1&-\frac25&\frac65&0\\ -2&1&3&1\\ \end{array}\right]$$

$$\left[\begin{array} {cccc} 1&-\frac25&\frac65&0\\ 0&\frac15&\frac{17}{5}&1\\ \end{array}\right]$$

$$\left[\begin{array} {cccc} 1&0&8&2\\ 0&1&17&5\\ \end{array}\right]$$

Am I screwing this up somewhere?

2. Oct 1, 2009

### Staff: Mentor

I ended up with different numbers in the 3rd column.
$$\left[\begin{array} {cccc} 1&0&12&2\\ 0&1&27&5\\ \end{array}\right]$$

In my first step, I added 2 times row 1 to -5 times row 2. Then I added 2 times row 2 to row 1. Finally, I replaced row 1 with 1/5 of itself. It is usually simpler to keep integer values for as long as possible.

3. Oct 1, 2009

### aPhilosopher

@Saladsamurai, [STRIKE]When you go to eliminate the -2, you want to add row 1 to row 2. You're subtracting it.[/STRIKE]

3 + 12/5 = 15/5 + 12/5 = 27/5

4. Oct 1, 2009