Talk:Reduced row echelon form: Difference between revisions
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=="Break" vs. "return" bug==
The algorithm has a Bug.<br>
It does NOT
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>
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