Applying Gaussian Elimination to a Matrix - Understanding the Correct Method

In summary, the conversation discusses Gaussian elimination on a matrix and the attempt at using a different method to solve it. The expert suggests that while the method is not illegal, it may not be the most efficient and recommends a simpler approach. The conversation ends with the understanding that the original method was incorrect and the expert provides the correct solution.
  • #1
robertjford80
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Homework Statement



Apply Gaussian elimination to the following matrix

2 -3 0 3
4 -5 1 7
2 -1 -3 5

I understand how to get the answer.

The Attempt at a Solution



My question is why doesn't the following method work: Switch Z over into the first column, move row 1 to row 3, then exchange rows 1 and 2 so that the new matrix is
1 -5 4 7
-3 -1 2 5
0 -3 2 3

Multiply the first row by 3 and add that to equation 2 to get:

1 -5 4 7
0 5 8 35
0 -3 2 3

Multiply row 2 by 3/5 and add that to equation 3 to get

1 -5 4 7
0 5 8 35
0 0 (34/5) 24

By this time I realized I was on the wrong track but I don't why what I did was illegal.
 
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  • #2
robertjford80 said:

Homework Statement



Apply Gaussian elimination to the following matrix

2 -3 0 3
4 -5 1 7
2 -1 -3 5

I understand how to get the answer.




The Attempt at a Solution



My question is why doesn't the following method work: Switch Z over into the first column, move row 1 to row 3, then exchange rows 1 and 2 so that the new matrix is
1 -5 4 7
-3 -1 2 5
0 -3 2 3
The 3rd row has a sign error.
robertjford80 said:
Multiply the first row by 3 and add that to equation 2 to get:

1 -5 4 7
0 5 8 35
0 -3 2 3
And you have some more error in the matrix above.
robertjford80 said:
Multiply row 2 by 3/5 and add that to equation 3 to get

1 -5 4 7
0 5 8 35
0 0 (34/5) 24

By this time I realized I was on the wrong track but I don't why what I did was illegal.
No, it's not illegal, but that's not to say it's a good idea.

The solution to the system that your augmented matrix represents is x = 3, y = 1, z = 0. If you switch columns you are in essence swapping variables.

When I did what you described, I quit at this matrix:
1 -5 4 | 7
0 8 -7 | -13
0 3 -2 | -3

I didn't bother to continue reducing it. Instead I checked that x = 0, y = 1, z = 3 was the solution, which it was.

BTW, you did an awful lot of extra work swapping columns and rows that you didn't really need to do. It's much easier to use the original row-1, col-1 number to eliminate the two entries under it, then use the row-2, col-2 entry to eliminate the entries above and below it. You quickly get to the solution this way.
 
  • #3
Good, thanks.
 

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