Matrices/Systems of Linear Equations

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DiamondV
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Homework Statement


Find the general solution:
http://puu.sh/ngck4/95470827b1.png

Homework Equations


Method: Gaussian Elimination by row operations.

The Attempt at a Solution


http://puu.sh/ngcml/7722bef842.jpg
I am getting the wrong answer( w = -27/5). The solutions provided to me says the answer is w=4. I double checked my row operation calculations(hence the ticks beside each operation) and I cannot seem to find my error.
 
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DiamondV said:

Homework Statement


Find the general solution:
http://puu.sh/ngck4/95470827b1.png

Homework Equations


Method: Gaussian Elimination by row operations.

The Attempt at a Solution


http://puu.sh/ngcml/7722bef842.jpg
I am getting the wrong answer( w = -27/5). The solutions provided to me says the answer is w=4. I double checked my row operation calculations(hence the ticks beside each operation) and I cannot seem to find my error.
Why is -10/6 equal to -1/10 in the operation ##r_3*1/6##?
 
Last edited by a moderator:
DiamondV said:

Homework Statement


Find the general solution:
http://puu.sh/ngck4/95470827b1.png

Homework Equations


Method: Gaussian Elimination by row operations.

The Attempt at a Solution


http://puu.sh/ngcml/7722bef842.jpg
I am getting the wrong answer( w = -27/5). The solutions provided to me says the answer is w=4. I double checked my row operation calculations(hence the ticks beside each operation) and I cannot seem to find my error.

It is a real waste of time to force the "diagonal" elements of your reduced form to all = 1; better to just leave them untouched. When I do it I obtain the final form as
[tex] \begin{array}{rrrr|r}<br /> 1 & 1& 1 & 1& 5 \\<br /> 0 & -2 &4 & 4 & 10 \\<br /> 0 & 0 & -2 & 0 &2\\<br /> 0 & 0 & 0 & -1 & -4<br /> \end{array}[/tex]
This is a triangular system, easily solved in the order ##w, z, y, x##. There is no good reason to divide all of rows 2 and 3 by -2 and row 4 by -1; the divisions can be done during the solution procedure. Of course, if you were taught to do it the way you did, then that is a different story. However, you still have errors somewhere, and I would need to see a typed version in order to take the time needed to find them.
 
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Samy_A said:
Why is -10/6 equal to -1/10 in the operation ##r_3*1/6##?
Ah. I got the right answer. I always make these mistakes in matrices. Its getting really annoying now. Thanks anyways.
 
Ray Vickson said:
It is a real waste of time to force the "diagonal" elements of your reduced form to all = 1; better to just leave them untouched. When I do it I obtain the final form as
[tex] \begin{array}{rrrr|r}<br /> 1 & 1& 1 & 1& 5 \\<br /> 0 & -2 &4 & 4 & 10 \\<br /> 0 & 0 & -2 & 0 &2\\<br /> 0 & 0 & 0 & -1 & -4<br /> \end{array}[/tex]
This is a triangular system, easily solved in the order ##w, z, y, x##. There is no good reason to divide all of rows 2 and 3 by -2 and row 4 by -1; the divisions can be done during the solution procedure. Of course, if you were taught to do it the way you did, then that is a different story. However, you still have errors somewhere, and I would need to see a typed version in order to take the time needed to find them.

I was taught to get all the leading entries of each row to 1 before solving it. But I did notice during one of the solutions to one of the questions provided to us, he solved a bit earlier when the leading entries were not 1. He said it was easier to do it now then later which matches up with what you're saying. But when exactly am I allowed to go straight to solving it and not leaving the leading entries as 1?
Also I don't understand where my other errors are? I got all the correct solutions for x,y,z,w using the fact that w=4