This matrix doesn't reduce!


by Pengwuino
Tags: matrix, reduce
Pengwuino
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#1
Feb16-06, 12:28 AM
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Ok am I missing something here or did our professor give us a take-home quiz with a question that has no answer? We needed to find the answer to a system of equations as follows:

4x-8y+3z=16
-x+2y-5z=-21
3x-6y+z=6

I setup the matrix….

[tex]
\begin{array}{*{20}c}
4 & { - 8} & 3 & {16} \\
{ - 1} & 2 & { - 5} & { - 21} \\
3 & { - 6} & 1 & 6 \\
\end{array}[/tex]

I reduced it down to

[tex]
\begin{array}{*{20}c}
0 & 0 & { - 17} & { - 68} \\
{ - 1} & 2 & { - 5} & { - 21} \\
0 & 0 & { - 14} & { - 57} \\
\end{array}[/tex]

At this point I realized… the equations didn't work. Am I right to think this quiz was given with a bad problem on it?

I even had mathematica reduce it and it gave me 2 rows that tried to say 1 = 0.
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Geekster
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#2
Feb16-06, 12:35 AM
P: 44
Quote Quote by Pengwuino
At this point I realizedů the equations didn't work. Am I right to think this quiz was given with a bad problem on it?
What's wrong with asking for a solution to a system with no solution? Your answer should be something like, "It's a singular matrix, no solution."
Pengwuino
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#3
Feb16-06, 12:46 AM
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Well i wouldn't expect something that easy on a quiz...

Geekster
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#4
Feb16-06, 12:49 AM
P: 44

This matrix doesn't reduce!


Quote Quote by Pengwuino
Well i wouldn't expect something that easy on a quiz...
So I guess that makes it a pretty good question for a quiz!
krab
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#5
Feb16-06, 01:21 AM
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The determinant of the matrix is zero, so there is no solution.
TD
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#6
Feb16-06, 05:07 AM
HW Helper
P: 1,024
A zero determinant of the coefficient matrix doesn't necessarily imply that there are no solutions. If the rank of the coefficient matrix is smaller than the rank of the 'extended matrix' (coefficient matrix + column of constants) then you have a system without a solution.
However, it's possible that although the determinant of the coefficient matrix is zero, the system still has solutions. You will find an infinite number of solutions in that case, the equations were then linearly dependant.


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