A TI-83 ? re MATRIX output (examples included)
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A TI-83 ? re MATRIX output (examples included)
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?
Help please?
curious angel
curious@sj.bigger.net
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