# Gaussian Elimination

Tags: elimination, gaussian
P: 35
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:

 [ 3 10 4 : 7 ] [ 2 7 3 : 5 ] [ 1 3 2 : 2 ]
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:

 [1 1 1 : 0] [3 4 8 : 1] [4 5 c^2 : c - 2]
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.
 PF Patron P: 1,235 $$\left(\begin{array}{x1x2x3k} 3 & 10 & 4 & 7\\ 2 & 7 & 3 & 5\\ 1 & 3 & 2 & 2 \end{array}\right)$$ Gaussian Elimination or any other fancy name is simply an attempt to reduce this matrix into a $$\left(\begin{array}{x1x2x3k} 1 & 0 & 0 & k1\\ 0 & 1 & 0 & k2\\ 0 & 0 & 1 & k3 \end{array}\right)$$ form. I will use notation R1, R2, and R3 to indicate Rows 1, 2, and 3. For example (R2 = R2 + R1 means you add row 1 into row 2, and after that you should have Row 1 as it was, and Row 2 will be sum of Row 1 and Row 2 from before. Now in your case: $$\left(\begin{array}{x1x2x3k} 3 & 10 & 4 & 7\\ 2 & 7 & 3 & 5\\ 1 & 3 & 2 & 2 \end{array}\right)$$ R1 = R1 - R2 -> $$\left(\begin{array}{x1x2x3k} 1 & 3 & 1 & 2\\ 2 & 7 & 3 & 5\\ 1 & 3 & 2 & 2 \end{array}\right)$$ R2 = R2 - 2*R3 -> $$\left(\begin{array}{x1x2x3k} 1 & 3 & 1 & 2\\ 0 & 1 & -1 & 1\\ 1 & 3 & 2 & 2 \end{array}\right)$$. R3 = R3 - R1 -> $$\left(\begin{array}{x1x2x3k} 1 & 3 & 1 & 2\\ 0 & 1 & -1 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$. $$\left(\begin{array}{x1x2x3k} 1 & 3 & 1 & 2\\ 0 & 1 & -1 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$ R1 = R1 - 3R2 -> $$\left(\begin{array}{x1x2x3k} 1 & 0 & 4 & -1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$. R1 = R1 - 4R3 -> $$\left(\begin{array}{x1x2x3k} 1 & 0 & 0 & -1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$. R2 = R2 + R3 -> $$\left(\begin{array}{x1x2x3k} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$ So you end up with this: $$x1 = -1, x2 = 1, x3 = 0$$ Your answer is: $$(-1, 1, 0) \epsilon \mathbb{R}^3$$

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