Re: A TI-83 ? re MATRIX output (examples included)
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Re: A TI-83 ? re MATRIX output (examples included)
here's what I did on 92:
[ 1 -1 1 4 ]
[ 5 2 -3 2 ]
[ 4 3 -4 -2 ] -->m
then I hit my fancy rref(m) and multiplied by (-)seven 'cause I don't
like fractions :-)
[-7 0 1 -10 ]
[0 -7 8 18 ] ingnore last row, make it parametric,
let z = t
now: t = 7x - 10
t = (7y + 18) / 8 {too hard
to type it out all steps, hope I did it right}
now you can graph, or at least I can on my 92 ;-)
you can also say:
7x-10 = (7y + 18) / 8 and solve
for y to put in "y=mx+b" form
************aka "eliminate the paramater"*********
note, when you have two or more parameters, or 3+ variables, it doesn't
make much sense to elimate parameter, but that's a whole 'nother can o'
worms beyond me.
maybe post your problem to ti-graph? lots of phd's over there, very
helpful.
BTW, usualy the "|" symbol means to take the determinant of matrix.
what class is this for?
hope that helps some,
Jesse1s@hotmail.com
you need to change the "c" to an "s" in my return address to foil spam
Curious Angel wrote:
> Hi:
>
> The problem I'm having is knowing how to read a matrix when the
> variables
> have not been isolated because the output is "Dependent" or "No
> Solution".
> I can quickly ascertain from the calculator if each variable has only
> one
> value that makes all linear equations true for that problem -- what if
> that's
> NOT the case, and you can't tell from the output what's what? Here's
> an
> example of what I mean:
>
> Take these 3 linear equations:
> x - y + z = 4
> 5x + 2y - 3z = 2
> 4x + 3y - 4z = -2
>
> The matrix is a 3 X 4 matrix, set up as follows:
> | 1 -1 1 4 |
> | 5 2 -3 2 |
> | 4 3 -4 -2 |
> it then produces:
> | 1 0 -1/7 10/7 |
> | 0 1 -8/7 -18/7 |
> | 0 0 0 0 |
> The answer to this particular set of equations is that they are
> "Dependent"
> (there is an infinite number of solutions for the x, y and z
> variables). But
> you can't instantly tell from the output that it IS Dependent, because
> along
> comes something like this, which has . . . "No Solution":
>
> x + z = 0
> x + y + 2z = 3
> y + z = 2
>
> We're missing a y and an x. Substituting zeroes for their
> placeholders, the
> matrix is again a 3 X 4 matrix, set up as follows:
> | 1 0 1 0 |
> | 1 1 2 3 |
> | 0 1 1 2 |
> and produces:
> | 1 0 1 0 |
> | 0 1 1 0 |
> | 0 0 0 1 |
>
> Thinking I've mastered how the TI-83 puts out, I write "Dependent" as
> my
> answer to this problem and promptly get it WRONG.
>
> Does the TI-83 require me to manually substitute in to solve? You
> must be
> kidding! How can I tell what I've got?
>
think this is special case, often if you have a square matrix you can do
augment and do gauss to find inverse, or use other techniques (many++),
I think once you have parameters you need to do it "manually". ??
> Help please?
> curious angel
> curious@sj.bigger.net
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