Mathematics
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OpenStudy (anonymous):
Linear algebra question?
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OpenStudy (anonymous):
OpenStudy (anonymous):
Anyone want to help me get started? I am pretty sure I can do the rest on my own.
OpenStudy (anonymous):
Represent the system in matrix form
OpenStudy (anonymous):
Yep, that's what I did.
OpenStudy (anonymous):
@Cecily
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OpenStudy (anonymous):
How do I go on from there?
OpenStudy (anonymous):
@zepdrix @Outkast3r09
OpenStudy (anonymous):
@dumbcow
OpenStudy (sirm3d):
take the determinant of the coefficient matrix, and set it equal to zero.
OpenStudy (anonymous):
Woah woah!!! We haven't learned what a determinant is yet. :P .
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OpenStudy (sirm3d):
okay.
how about gauss-jordan reduction?
OpenStudy (anonymous):
Yeah. We learned that.
OpenStudy (sirm3d):
i got something easier than gauss-jordan.
multiply equation 2 by (-1), and equation 3 by 2, then add the two equations.
OpenStudy (anonymous):
Elimination?
OpenStudy (sirm3d):
yup.
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OpenStudy (anonymous):
3y-3z=-a+2 ?
OpenStudy (sirm3d):
not 3y+3z = -a + 2 ?
OpenStudy (anonymous):
Yeah. My mistake.
OpenStudy (anonymous):
Allright, what next?
OpenStudy (sirm3d):
use that equation with equation 1 to eliminate both y and z.
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OpenStudy (anonymous):
1 sec.
OpenStudy (anonymous):
THat dosen't eliminate it. It makes it bigger.
OpenStudy (sirm3d):
bx + 3y + 3z = a
3y + 3z = 2 - a
subtract.
OpenStudy (anonymous):
I used substitution instead. Should till be valid right?
OpenStudy (sirm3d):
or if you wish to substitute...
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OpenStudy (sirm3d):
that's valid too.
OpenStudy (sirm3d):
bx + (2-a) =a
bx = 2a - 2
OpenStudy (anonymous):
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