ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: Odette on February 28, 2008, 06:53:08 pm

Title: Help with Matrices
Post by: Odette on February 28, 2008, 06:53:08 pm

How do you get from
   This
[2 1 3 -5]
[3 5 6 -5]
[4 8 8 -5]
     to
[1   2    2    -1.25]
[0   1 (1/3)  (5/6)]
[0   0    1    -1.25]
    That ?

Your help will be appreciated. Thanks (Btw its to row echelon form) 
 

Title: Re: Help with Matrices
Post by: cosec(x) on February 28, 2008, 07:34:01 pm

Is it an augmented matrix used to solve three simultaneous equations
By using:
2x+y+3z=-5    (1)
3x+5y+6z=-5   (2)
4x+8y+8z=-5  (3)

I got
(3)  - 2(1)
6y+2z=5  (4)

2(2) - 3(1)
7y+3z=5   (5)

From this
2(5) - 3(4)
-4y=-5
therefore y=1.25
from this we can see z=-5/4, x= 5/8 (by subbing in values into prev. equs)

Alternatively, matrices can be used to do this (the purpose of reduced row echelon form)

My reduced matrix was

[2 1  3 | -5]
[0 7  3 | 5 ]
[0 0 -4 | 5]

(obtained by step1: 2r2-3r1; r3-2r1  step2: 7r3-6r1)
It strikes me as odd that the first row is different in your desired result, as in making reduced row echelon form, you don't need to touch the top row

hope it helps!!!!

Title: Re: Help with Matrices
Post by: midas_touch on February 28, 2008, 08:38:54 pm

There is probably a more efficient method in doing this, but heres a series of steps i used to get the row echelon form of the above matrix:

[2 1 3] -5
[3 5 6] -5
[4 8 8] -5

Divide row 3 by 4, and swap with row 1

[1 2 2] -5/4
[3 5 6] -5
[2 1 3] -5

Subtract row 2 by 3 times row 1, and subtract row 3 by 2 times row 1

[1 2 2] -5/4
[0 -1 0] -5/4
[0 -3 -1] -5/2

Multiply row 2 by negative 1, divide row 3 by negative 3, then swap rows 2 and 3

[1 2 2] -5/4
[0 1 1/3] 5/6
[0 1 0] 5/4

Subtract row 3 by row 2

[1 2 2] -5/4
[0 1 1/3] 5/6
[0 0 -1/3] 5/12

Finally, multiply row 3 by -3

[1 2 2] -5/4
[0 1 1/3] 5/6
[0 0 1] -5/4

Also, you shouldn't work with decimals and fractions in the same problem, choose either one or the other

Title: Re: Help with Matrices
Post by: Odette on February 28, 2008, 09:54:52 pm

Btw how did you get 5/8 for x? i keep getting -5/4 for x :(

Title: Re: Help with Matrices
Post by: cosec(x) on February 28, 2008, 10:02:08 pm

sorry, its meant to be x=-5/4 typo :-[

drat, this is what happens when you do your working on the back of an envelope :uglystupid2:

Title: Re: Help with Matrices
Post by: Odette on February 28, 2008, 10:09:10 pm

Oh that's ok lol thanks for your help ! I just thought I was doing something wrong >.<