Re: A TI-83 ? re MATRIX output (examples included)
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Re: A TI-83 ? re MATRIX output (examples included)
Curious Angel wrote:
> 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".
........
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).
"Dependent" is because the last equation has vanished altogether (become
0 = 0). Whether there are no solutions, or infinitely many solutions,
or just one solution, depends on what comes above the zero row. In this
case there are infinitely many solutions because you have only two
useful equations in three 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 |
This time the equations are not dependent, because no row has vanished;
on the other hand the last equation now says 0 = 1, which is
impossible, so there are no solutions. [No combination of numbers makes
0 = 1, so no combination of numbers makes the reduced equations true, so
no combination of numbers makes the original equations true.]
> 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?
With three equations in three unknowns, the most likely outcomes are:
Last row 0 0 0 a : no solutions
Last row 0 0 0 0 : infinitely many solutions
Last row 0 0 a b : unique solution
[Other outcomes are possible.]
In general for m equation in n unknowns, if you EVER get a row that
says 0 0 0 .... 0 a , for a * 0, stop, because there are no solutions.
Otherwise if there are as many non-zero rows left as there are
variables, you get a unique solution;
If there are fewer non-zero rows than variables, there are infinitely
many solutions.
Whether the equations were dependent or not is a different question. If
you get a row entirely zero they were; but there still might be one, or
many, or no solutions.
P.S (and N.B.) I've decided to unsubscribe from CALC-TI, for the VERY
obvious reason, but I shall remain on GRAPH-TI. Please send
mathematical questions to GRAPH-TI rather than CALC-TI, which is for a
different sort of user.
Chris Barling
Swinburne University of Technology
Melbourne, Australia
References: