Gaussian Elimination - Help

MR DAZ

Gaussian Elimination HELP!!!!
Lads, I'm studying for a diploma in IT through dstance learning(oscial)......and the old exmas are coming up.....

I was working my way through this years sample paper and...I get this one to work out.....I know that sometimes they dont work out...but I'm looking for some reassurance/help

gaussian Elimination:

a + 3b - 2c + 3d = 2
4a + 12b -8c +14d = 6
-2a -6b + 4c +d = -11
3a + 9b - 6c + 5d = 10

I cant seem to get it to work out....would really appreciate any help on this.



thanks again

jd

where did you get stuck..?
lets see your augmented matrix>?

ecksor

Gauss-Jordan is the same method as row-reduction, but without the augmented matrix. I.E, you keep subtracting as you would with an augmented matrix to get leading 'ones', or (put another way) consecutive expressions with one less variable each time. The elementary row operations are the same. (I'm used to the augmented matrix notation too, had to check a text book to remember the difference).

Anyway, to answer the original question, it's basically an algorithm that specifies how to solve a system of linear equations, and it very probably mimics the natural way of doing things you would have had before with systems of 2 or 3 equations, but tidies things up slightly. The elementary row operations are:

(1) - Swap rows (equations) around.
(2) - Subtract a multiple of one equation from another.
(3) - Multiply an equation by an arbitrary value.

Using those, you aim to end up with something that looks like:

a            = *
  b          = *
    c        = *
      d      = *

(with numbers instead of stars obviously ) from which you can read off your answer.

So, do everything you need to do to get one equation in terms of an expression with a leading a. Once that is done, you can keep subtracting multiples of that equation until you have another equation with no a and a leading b. Use those two to remove the a and b from the next and multiply to get the leading c etc ...

One you have all leading variables with a co-efficient of 1, you can work backwards to remove the trailing variables from the leading edge. (i.e, a multiple of the equation d = * can be used to eliminate the d from all other equations, etc).

MR DAZ

right this is how far i make it.....

3 + 9 -6 + 5 10
-8 - 24 +16 -12 -28
0 0 -32 -16 16
0 0 0 -2 2

try and finish it of and see what happens......

logic1

Here's a handy little tool to help you out:

http://www.math.ucla.edu/~tao/115a.3.02f/Gauss.html

.logic.

ecksor

Originally posted by MR DAZ
try and finish it of and see what happens......

You won't necessarily come up with a unique solution. Sometimes there are no solutions, and sometimes there are an infinte number of solutions that obey an equation.

In this particular system of equations, I reduced the last 3 by adding the appropriate multiple of the first row (purely because it seemed like the simplist to work with), and the last 3 rows all ended up being in terms of d but I got a consistent value for d, so substitution of that into the first equation gives an answer in terms of a, b and c.

If there wasn't a consistent value for d when you get that sort of thing then the equations would have no solution.

a + 3b - 2c = 5, d = -1

MR DAZ

Any chance you might post your method....

I know how to do the equation/work them out

but for this particular one...it doesnt work out! or else i'm doing it wrong, which in that case show me my mistake please

GuanYin

I worked this out as far as is possible last night coz I was bored.

I hate duplicate threads though.

Capt'n Midnight

Since all 4 quadratics contain the term n(a + 3b) you can't solve for a or b - but as long as c and d are consistant you should end up with an answer in the form of A + mB + const = 0 - ie. a straight line..

cf. http://www.boards.ie/vbulletin/showthread.php?postid=1169772#post1169772

ecksor

n(a + 3b - 2c) I think you mean.

ecksor

Originally posted by MR DAZ
Any chance you might post your method....

I did! Twice!!

  a +  3b - 2c +  3d = 2   (r1)
 4a + 12b - 8c + 14d = 6   (r2)
-2a -  6b + 4c +   d = -11 (r3)
 3a +  9b - 6c +  5d = 10  (r4)

a + 3b - 2c + 3d =  2 (r1)
              2d = -2 (r2 - 4r1)
              7d = -7 (r3 - 2r1)
             -4d =  4 (r4 - 3r1)

At which point I concluded that d = -1 and substituted back into r1, so:

a + 3b - 2c -3 = 2

a + 3b - 2c    = 5

MR DAZ

Thanks, lads for all your Help... really appreciate it!

especially Ecksor for typing everything out.

you can see that from my answer posted above that I was getting d = -1 aswell, maybe i didnt state it right but all i wanted to know was this as far as you could go with the guassian elimination.

Thanks for for clarifying it guys!!!!!

Capt'n Midnight

n(a + 3b - 2c) I think you mean.

No - 'cos in the third equation the sign for C has changed

ecksor

Originally posted by Capt'n Midnight
n(a + 3b - 2c) I think you mean.

No - 'cos in the third equation the sign for C has changed

n = -2

Capt'n Midnight

Ah - it was a Typo when I popped into this spreadsheet..