Talk:Reduced row echelon form: Difference between revisions

Content added Content deleted
No edit summary
No edit summary
Line 1: Line 1:
=="Break" vs. "return" bug==
=="Break" vs. "return" bug==


The algorithm has a Bug.
The algorithm has a Bug.<br>
It does NOT do the SWAP. No conditions are checked


It does NOT work for this example matric
It does NOT do the SWAP. No conditions are checked<br>


Solve the following system of equations.
3x+y− 4z=−1
x +10z= 5
4x+y+ 6z= 1
Solution. The corresponding augmented matrix is
3 1 −4 −1
1 0 10 5
4 1 6 1
Create the first leading one by interchanging rows 1 and 2
1 0 10 5
3 1 −4 −1
4 1 6 1
Now subtract 3 times row 1 from row 2, and subtract 4 times row 1 from row 3. The result is
1 0 10 5
0 1 −34 −16
0 1 −34 −19
Now subtract row 2 from row 3 to obtain
1 0 10 5
0 1 −34 −16
0 0 0 −3
This means that the following reduced system of equations
x +10z= 5
y−34z=−16
0= −3


It does NOT work for this example matrix<br>
is equivalent to the original system. In other words, the two have the same solutions. But this last

system clearly has no solution (the last equation requires that x, y and z satisfy 0x+0y+0z = −3,

and no such numbers exist). Hence the original system has no solution.
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 10 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= −3<br>


is equivalent to the original system. In other words, the two have the same solutions. But this last<br>

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). Hence the original system has no solution.<br>