Need help with a Matrix/Linear Algebra Problem

VulnoX · June 2007

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!

Drakmathus · June 2007

3 equations, thre unknowns. bust it out with the solve function on your calculator.

Serpent · June 2007

uh

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.

VulnoX · June 2007

Drakmathus wrote: »

3 equations, thre unknowns. bust it out with the solve function on your calculator.

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.

VulnoX · June 2007

Serpent wrote: »

uh

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.

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.

QuasiStatic · June 2007

no it doesn't, nevermind.

if all else fails just be a sly bastard and use the zero vector.

Drakmathus · June 2007

hit 2nd then home (custom).

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.

VulnoX · June 2007

Drakmathus wrote: »

hit 2nd then home (custom).

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.

QuasiStatic · June 2007

VulnoX wrote: »

Drakmathus wrote: »

hit 2nd then home (custom).

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

VulnoX · June 2007

I typed it in, and it says:
Error: Argument must be a Boolean Expression, I am trying some different stuff, if you know a reason for this it would be appreciated.

Serpent · June 2007

QuasiStatic wrote: »

x3 must equal 0.

false!

but it is a bit of a trick question, if that helps the OP at all.

a calculator is not needed at all

Drakmathus · June 2007

did you put the equal signs in? using X,Y,Z?

VulnoX · June 2007

Serpent wrote: »

QuasiStatic wrote: »

x3 must equal 0.

false!

but it is a bit of a trick question, if that helps the OP at all.
a calculator is not needed at all

If not, then that really sucks, I have been staring at this part since yesterday.

QuasiStatic · June 2007

U N S O L V A B L E

The last row is all 0s after a downward elimination.

???

VulnoX · June 2007

Drakmathus wrote: »

did you put the equal signs in? using X,Y,Z?

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?

Serpent · June 2007

the answer is a line itself.

QuasiStatic · June 2007

| -.4 .8 .7|
| .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

VulnoX · June 2007

HA!

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.

Serpent · June 2007

z=1 is not the correct answer. the answer is a line, not a point.

QuasiStatic · June 2007

Damnit how do you do this by HAND

VulnoX · June 2007

Serpent wrote: »

z=1 is not the correct answer. the answer is a line, not a point.

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.

VulnoX · June 2007

QuasiStatic wrote: »

Damnit how do you do this by HAND

Trust me, I wish I knew. Not all problems can easily be done by hand. I am sure its possible though.

Serpent · June 2007

the answer is a line. that's why your calculator gave you what it did. It's also what I got by hand, which is the same way quasistatic did it.

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?

QuasiStatic · June 2007

That did not seem intuitive for some reason

TechBoy · June 2007

VulnoX wrote: »

Serpent wrote: »

z=1 is not the correct answer. the answer is a line, not a point.

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.

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.

Odinious · June 2007

If you are using some sort of TI calculator, you can use the reduced row echelon function to find the solution. Put the matrix into your calculator using the following convention...

[HTML]

[FONT="Courier New"]
A system of equations...

-.4x + .8y + .7z = 0
 .2x - .9y + .1z = 0
 .2x + .1y - .8z = 0

is entered into a 3x4 matrix like TechBoy posted...

| -.4  .8  .7 | 0 |
|  .2 -.9  .1 | 0 |
|  .2  .1 -.8 | 0 |

Exit back to the calculation screen, and look up the reduced row echelon function rref( under the "math" menu in the "matrix" menu and insert the matrix letter into the function. Close it out, and it gives you...

| 1 0 -3.55 | 0 |
| 0 1  -0.9 | 0 |
| 0 0     0 | 0 |

Which can be converted back into a system of equations...

1x + 0y - 3.55z = 0
0x +1y -   0.9z = 0

[/FONT]

[/HTML]

That being said, it doesn't have one specific solution.

VoodooV · June 2007

man, its so depressing that just a year ago, I knew how to do that, and now its almost completely forgotten

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

Penny Arcade

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