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sydboe · Mar 21, 2009, 10:28 AM

need work please!

x+2y+2z-w=0
x-y+3w=7
2x-y+z+w=6
2x+2y+3z-3w=-1

Curlyben · Mar 21, 2009, 10:54 AM

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galactus · Mar 21, 2009, 12:02 PM

I would solve this using elimination via Gaussian elimination.

Turn it into a matrix and reduce it using reduced row echelon form.

$\begin{bmatrix}1&2&1&-1&0\\1&-1&3&0&7\\2&-1&1&1&6\\2&2&3&-3&-1\end{bmatrix}$

When you get it into this form, you have your solutions:

$\begin{bmatrix}1&0&0&0&x\\0&1&0&0&y\\0&0&1&0&z\\0& 0&0&1&w\end{bmatrix}$

Your solutions are in the right most column.

sydboe · Mar 22, 2009, 07:16 PM

I'm totally stuck. I've been trying to cancel out things and just wind up cancelling out the whole equation, unable to find what any letters equal

galactus · Mar 23, 2009, 10:40 AM

This is tedious stuff. In my opinion, these should be done with a calculator.

I am sorry, but I am not going to post all this step by step.

Here is a screen capture from my TI-92. Now, at least, you can see what you're shooting for.

tranced · Apr 7, 2010, 11:09 PM

the answer is
x = 16
y = 9
z =-17
w = 0 and the row manipulation is as follows

r2-r1
r3-r4
r4-2r1
r2-2r4
r3+3r2
r4+2r2
r3/-2
r4+r3
r1-2r2
r1-2r3
r1-r4
r3+2r4

tranced · Apr 7, 2010, 11:10 PM

By the way I used gauss jordan

ebaines · Apr 8, 2010, 10:36 AM

Originally Posted by tranced

btw i used gauss jordan

Let's just hope the OP hasn't been waiting around for over a year to turn in his homework!