Simple Gauss-Jordan Elimination

[Saladsamurai]

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:

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

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

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

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

Am I screwing this up somewhere?

 

[Mark44]

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:

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

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

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

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

Am I screwing this up somewhere?

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

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.

 

[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

 

[Saladsamurai]

Yeah! I just caught that :/ Man.. I must have added 6/5+3=17/5
like ten times! Stupid brain gets sucked into following
a pattern

thanks!