Reduced row echelon form: Difference between revisions

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Hello RosettaCode ...<br>
 
Your algorithm/function does NOT work ...<br>
 
The "SWAP" is not being done. ...<br>
 
Try this matrix to see the problem. ...<br>
 
 
//===================================<br>
 
Linear Algebra with Applications by W. Keith Nicholson, University of Calgary<br>
 
page 12 - Systems of Linear Equations<br>
 
 
Solve the following system of equations.<br>
 
3x + y − 4z = −1<br>
 
x + 10z = 5<br>
 
4x + y + 6z = 1<br>
 
 
Solution. The corresponding augmented matrix is<br>
 
3 1 −4 −1<br>
 
1 0 10 5<br>
 
4 1 6 1<br>
 
 
Create the first leading one by interchanging rows 1 and 2<br>
 
1 0 01 5<br>
 
3 1 −4 −1<br>
 
4 1 6 1<br>
 
 
Now subtract 3 times row 1 from row 2, and subtract 4 times row 1 from row 3. The result is<br>
 
1 0 10 5<br>
 
0 1 −34 −16<br>
 
0 1 −34 −19<br>
 
 
Now subtract row 2 from row 3 to obtain<br>
 
1 0 10 5<br>
 
0 1 −34 −16<br>
 
0 0 0 −3<br>
 
 
This means that the following reduced system of equations<br>
 
x + 10z = 5<br>
 
y − 34z = −16<br>
 
0 +0 +0 = −3<br>
 
 
is equivalent to the original system. In other words, the two have the same solutions. <br>
 
But this last system clearly has no solution (the last equation requires that x, y and z satisfy 0x+0y+0z = −3,<br>
 
and no such numbers exist). <br>
 
Hence the original system has no solution.<big>Big text</big>
 
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{{wikipedia|Rref#Pseudocode}}
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