Gaussian Elimination

adc85

I tried to solve this Gaussian elimination algorithm problem (matrices) but for some reason when I plug in the x variables it doesn't work. The problem is:

Alright so the first thing I did was divide the 1st row by 1/3 (scaling). Then I made the entries below the first pivot equal to 0 using:

Row2 = Row2 - Second Row, First Column * Row 1
Row3 = Row3 = Third Row, First Column * Row 1

Then I repeat this algorithm for the submatrix created afterwards (ignoring the first row and first column). Afterwards, I used backwards substitution (even tried using reduced echelon form). But I am not getting quite the right answers (very close for row 3 and the other two rows are fine though). Any input appreciated.

Also, I am having trouble understanding what my professor is saying when he says to solve a certain problem like this using 10^-3 precision for example. Do you just use the same method except placing decimal places at the end of each number or whatever?

Then he has this other weird problem that goes like:

What value of c would make this inconsistent (in other words, no solution)? I'm thinking that x3 could equal anything by making the last row full of zeros. Not sure though.

Thanks for any help.

gnome

As to your first question, can't tell you where you're going wrong if you don't show your work. Just a thought though: rather than starting off dividing (and immediately dealing with fractions) why not just swap the 1st & 3rd rows?

As to the second, I don't understand what you're asking.

As to the third, if you end up with all zeroes in the last row (i.e. by making c=3), you end up with a free variable, NOT an inconsistent system. So keep thinking about what value you can give to c, so you end up with zeroes in the first 3 columns but a non-zero value in the last column. THAT will be a system with no solution.

adc85

Thank you gnome.

So if any row has this set-up:

[ 0 0 0 : Non-Zero Number ]

Then it means that it has no solution? Even if it's the second row?

gnome

That's right.