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Mathematics 11 Online
OpenStudy (anonymous):

Linear algebra question?

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

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 .

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.

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.

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...

OpenStudy (sirm3d):

that's valid too.

OpenStudy (sirm3d):

bx + (2-a) =a bx = 2a - 2

OpenStudy (anonymous):

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