Re: Help 3 vars. equ's
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aaron wrote:
>
> My math teacher said that there was a way to use the matrices to solve a
> system of 3 equations with 3 differnt variables. I have a TI-85
> Does anyone know how to do it????
> please help i hate writing all that stuff out!! hehe :)
>
> Thanks
Yep.. There are two ways: you can use the built-in matrix editor, or you
can just type the matrix out on the home screen. I'm going to show you
how to do it on the home screen. If you want to know how the built-in
editor works, just play with it (or, alternatively, read the manual).
Say you've got these three equations: 5x+6y+2z=0, 3x+4y+9z=5, 3y+7z+x=6.
First, rewrite them so the terms are in the same order (first the x's,
then y's, then z's):
5x +6y +2z =0
3x +4y +9z =5
1x +3y +7z =6
Now we'll build the matrix in the calc. TI-85 matricies use the square
brackets around each row, with commas between each element. When writing
the matrix, we only use the coeffecients of each term, so here goes:
[[5, 6, 2, 0]
[3, 4, 9, 5]
[1, 3, 7, 6]]
When you type it in, it can all go on one line. The spaces aren't needed
in the calc, either. If you hit the STO> key and give a name, then the
matrix will be stored in memory. Now, to put the matrix in reduced-row
echelon form (simplify):
rref <matrix>
You could either actually type the matrix in after the rref, or you
could store the matrix into a variable, for example M, and then do rref
M. What you'll get back:
[[1 0 0 -1.684...]
[0 1 0 1.2105...]
[0 0 1 .57894...]]
This could be made back into our three equations with three variables
as:
1x +0y +0z =-1.684 =x
0x +1y +0z =1.2105 =y
0x +0y +1z =.57894 =z
There, now your equations are simplified. Wasn't that easier than
substitution and algebra? :)
<pre>
--
Greg Hill
http://braves.bhs.davis.k12.ut.us/
ghill@lab.tech.bhs.davis.k12.ut.us
greg_hill@email.state.ut.us
</pre>
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