Solve System of Equations Using Gaussian Elimination

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The discussion focuses on solving a system of equations using Gaussian elimination. The initial equations are rearranged for better pivoting, and the process involves calculating new rows based on pivot factors. A participant identifies potential arithmetic mistakes in the calculations, particularly in the signs during the elimination steps. The importance of using the extended matrix form for clarity is also mentioned. The thread emphasizes the need for careful arithmetic to avoid errors in the final solution.
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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 that's wrong...can anyone point out where I've went wrong?
 
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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}<br /> 2 &amp; 1 &amp; { - 3} &amp;\vline &amp; 3 \\<br /> 4 &amp; { - 1} &amp; 2 &amp;\vline &amp; {25} \\<br /> { - 3} &amp; 2 &amp; 4 &amp;\vline &amp; { - 9} \\<br /> <br /> \end{array} } \right)
 
Bucky said:
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
presumably you meant here "R3- (1/2)R1" but the calculation is correct.

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)
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.

= 0y + ((33/6)z - (20/6)z) = 58.5/6 - 47.5/6
= (13/6)z = 11/6
z = 11/13
and i know that's wrong...can anyone point out where I've went wrong?
 
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