Expert: Paul Klarreich - 2/19/2009

Question
I just don't know how to solve this,the question states that use the gaussian elimination algorithm to solve the following linear systems,if possible,and determine whether row interchanges are necessary:
x1+x2   +x4=2,
2x1+x2-x3+x4=1,
4x1-x2-2x3+2x4=0,
3x1-x2-x3+2x4=-3.

Answer
Questioner:   Andrew
Category:  Calculus
Private:  Yes
 
Subject:  Gaussian elimination algorithm of linear systems of equation
Question:  I just don't know how to solve this,the question states that use the gaussian elimination algorithm to solve the following linear systems,if possible,and determine whether row interchanges are necessary:



x1  +  x2        +  x4 = 2,
2x1 +  x2 -  x3  +  x4 = 1,
4x1 -  x2 - 2x3  + 2x4 = 0,
3x1 -  x2 -  x3  + 2x4  =-3.


Start by getting rid of those wasteful symbols and write:


1    1    0     1  =   2
2    1   -1     1  =   1
4   -1   -2     2  =   0
3   -1  - 1     2  =  -3

Now we (that means 'you') can get to work.  Your ultimate objective is to get a 'table' that looks like this:


1   ?   ?   ?  = ?
0   1   ?   ?  = ?
0   0   1   ?  = ?
0   0   0   1  = ?

When that is achieved, you have your solution.  Effectively, the last line gives you a value of x4.  Now you can either do ordinary back-substitution, or do some more elimination to put 0's in place of the '?' on the left and get the rest of the solution.

Now to get started, note that the first line starts with '1'.  That is very nice.  use it to eliminate the 2,3,4:


1    1    0     1  =   2
2    1   -1     1  =   1  < Row 1 * 2 and subtract:
4   -1   -2     2  =   0  < Row 1 * 4 and subtract:
3   -1  - 1     2  =  -3  < Row 1 * 3 and subtract:



1    1    0     1  =   2
0   -1   -1    -1  =  -3
0   -5   -2    -2  =  -8
0   -4  - 1    -1  =  -9


OK, now I would change the signs in the second row and proceed from there downward -- killing the -5,-4 rows.  On down, and you will get it.
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