# Gaussian Elemination

1. Oct 24, 2005

### Bucky

use gaussian elimination to solve the following system of equation:

2x + y - 3z = 3
4x - y + 2z = 25
-3x + 2y + 4z = -9

ok so first i rearranged them so the inital numbers were in decending order (for the pivot)

R1 -> 4x - y + 2z = 25
R2 -> -3x + 2y + 4z = -9
R3 -> 2x + y - 3z = 3

pivot factor r = -3/4 for R2
pivot factor r = 1/2 for R3

R1 = 4x - y + 2z = 25
R2 + (3/4)R1 = (-3x+(3/4)4x) + (2y + (-3/4y)) + (4z + (3/4)2z) = -9 + (3/4)25
= 0x + 1.25y + 5.5z = 9.75

R3 = (2x - 1/2x) + (y +1/2y) + (-3z-z) = 3 - 25/2
= 0x + 1.5y - 4z = -9.5

now solve for R2 and R3 to find Z

pivoting factor r = 1.25/1.5 = 5/6

R2 - 5/6R3 = (1.25y - 5/6(1.5y)) + (5.5z - 5/6(4z)) = 9.75 - 5/6(9.5)
= 0y + ((33/6)z - (20/6)z) = 58.5/6 - 47.5/6
= (13/6)z = 11/6
z = 11/13

and i know thats wrong...can anyone point out where i've went wrong?

2. Oct 25, 2005

### TD

You probably made some arithmetic mistakes, but I didn't check it all.
Personally, when using Gaussian elimination, I'd use the extended matrix form and just work with the coëfficiënts.

Try to put the following matrix in row-reduced form (if you've seen that of course) and 'translate' back into equations.

$$\left( {\begin{array}{*{20}c} 2 & 1 & { - 3} &\vline & 3 \\ 4 & { - 1} & 2 &\vline & {25} \\ { - 3} & 2 & 4 &\vline & { - 9} \\ \end{array} } \right)$$

3. Oct 25, 2005

### HallsofIvy

Staff Emeritus
presumably you meant here "R3- (1/2)R1" but the calculation is correct.

Here's you're problem! Your signs are wrong on the last two terms.
R3 is 1.5- 4x= -9.5 so R2-(5/6)R3 is
(1.25y- (5/6)(1.5y))+ (5.5z-5/6(-4z))= 9.75- (5/6)(-9.5)
(5.5+ 20/3)z= 9.75+ 95/12.