Applying Gaussian Elimination to a Matrix - Understanding the Correct Method

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The discussion focuses on the application of Gaussian elimination to a specific matrix and addresses a user's confusion regarding the legality of their method. The user attempted to swap rows and columns but was informed that while not illegal, it was unnecessarily complicated and could lead to errors. The correct approach involves using the original matrix to eliminate entries systematically without unnecessary row or column exchanges. The solution to the system represented by the matrix is confirmed as x = 3, y = 1, z = 0. The conversation emphasizes the importance of following a straightforward elimination process for clarity and accuracy.
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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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.
 
Good, thanks.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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