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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