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Need help with a Matrix/Linear Algebra Problem
VulnoX
Registered User regular
Registered User regular
Ok, this is probably going to have a simple solution, but I have always had trouble with these kinds of things and need it for an assignment.
Basically, I have a matrix, say it has the values:
| -.4 .8 .7|
| .2 -.9 .1|
| .2 .1 -.8|
Now I need to find values for x1, x2, and x3 that are true for each row that makes it equal zero.
so,
-.4x1+.8x2+.7x3=0
.2x1-.9x2+.1x3=0
.2x1+.1x2-.8x3=0
So x1, x2, and x3 has to work for each problem.
Another example of what I mean.
If you had:
-2x1+4x2=0
2x1-4x2=0
x1 would equal 2 and x2 would equal 1 since those are the x values that work for both equations to make it zero.
I am trying to find an easy way, maybe with the calculator or something that will give me x1, x2, and x3. If anyone knows, I would greatly appreciate the help.
Thanks!
Basically, I have a matrix, say it has the values:
| -.4 .8 .7|
| .2 -.9 .1|
| .2 .1 -.8|
Now I need to find values for x1, x2, and x3 that are true for each row that makes it equal zero.
so,
-.4x1+.8x2+.7x3=0
.2x1-.9x2+.1x3=0
.2x1+.1x2-.8x3=0
So x1, x2, and x3 has to work for each problem.
Another example of what I mean.
If you had:
-2x1+4x2=0
2x1-4x2=0
x1 would equal 2 and x2 would equal 1 since those are the x values that work for both equations to make it zero.
I am trying to find an easy way, maybe with the calculator or something that will give me x1, x2, and x3. If anyone knows, I would greatly appreciate the help.
Thanks!
VulnoX on
0
Posts
you know
a big advantage of matrices is to do the exact problem easily by hand? matrices are setup to allow for extremly easy solving of multiple linear equations. using a calculator defeats the purpose.
Sorry, I have never used the solver before.
I have a TI-89 Titanium, and I went to the equation solver. I put it in as -.4x+.8y+.7z and it returned:
eqn:=
x:=
y:=
z:=
bound=(-1.E14,1.E14)
I am probably just using it incorectly, I am going to research using it a bit more.
But also remember that the X,Y, and Z values have to hold true for all three equations, thats where my problem is.
I have done the whole thing by hand, but finding those values is close to impossible without a calculator, which are my professors exact words.
My problem is not with a matrix, it is with finding variables, as I clearly stated.
if all else fails just be a sly bastard and use the zero vector.
then hit f3, 1 (solve( ). type in the first equation, then hit f3 again and then hit 2 (and), tpye in the second, f3 2 to get and again. type in the final equation then hit comma, x and close all parentheses you have (should only need 1). hit enter and assuming you havent typed in anything wrong, itll spit out all 3 values.
edit: hit 2nd home again to get back to the regular menu.
I am going to try this now, thank you very much for taking the time to type that out.
post answer plz
Error: Argument must be a Boolean Expression, I am trying some different stuff, if you know a reason for this it would be appreciated.
false!
but it is a bit of a trick question, if that helps the OP at all.
If not, then that really sucks, I have been staring at this part since yesterday.
The last row is all 0s after a downward elimination.
???
That was the problem, I thought about it while typing it in but then left it out.
It returned:
x=3.55 * z and y = .9 * z
I am guessing z is supposed to be 1?
| .2 -.9 .1|
| .2 .1 -.8|
| -.4 .8 .7|
| 0 -.5 .45|
| .2 .1 -.8|
| -.4 .8 .7|
| 0 -.5 .45|
| 0 .5 -.45|
| -.4 .8 .7|
| 0 -.5 .45|
| 0 0 0|
not solvable
***actually this is wrong, because the last line is actually |0 0 0| = 0, which is legal.
jousting at windmills a bit
That worked, so x=3.55, y=.9, and z=1
Thank you so much for your help, I need to remember this for the future, finding variables has always been my weakest point in math.
Thank you again.
Well it works in all problems giving me 0, and the answer can't be a line because this is for finding long term results from a Markov chain.
Trust me, I wish I knew. Not all problems can easily be done by hand. I am sure its possible though.
the answer (x = 3.55z, y = 0.9z) is just a line. ANY value of z will work here. z=0, then x=0 and y=0. z=2, then x = 7.1 and y = 1.8. get it?
You take the coefficient matrix
| -.4 .8 .7|
| .2 -.9 .1|
| .2 .1 -.8|
and append the solutions of the equations. In this case they're all zero, but they don't have to be. Thus creating
| -.4 .8 .7 | 0 |
| .2 -.9 .1 | 0 |
| .2 .1 -.8 | 0 |
You then row reduce the coefficient matrix with Gauss-Jordan elimination, making sure that any operation you do to the coefficient matrix row, you do to the corresponding solution.
Example here, http://ascworkshop.info/AiS/textbook/unit2/example_projects/starter/math/matrix/gauss.html
Serpent is also right about it being a line.
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That being said, it doesn't have one specific solution.
But I have to agree with Techboy and Serpent, pretty much all of my Matrix Theory class was doable without a calculator. If anything it took me more time to do a problem with a calculator since I would always forget how to enter everything or I couldn't remember where the particular function I wanted was in the menu. Even then, the only thing the calculator was useful for (and the only thing we were allowed to use it for) was to find the REF or the RREF
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