1 0 0 K1

0 1 0 k2

0 0 1 k3

3.) x - 3y -7z = 6

4x + y = 7

2x + 3y +z = 9

4.) 3x - 5z = -1

2x + 7y = 6

x + y + z = 5

... Help T_T this is trial & error probs ... so many, so complicated ... kindly teach me some tips in solving matrices?

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- Sep 9th 2006, 02:04 AM^_^Engineer_Adam^_^Kindly Help me plss!!! Matrices
1 0 0 K1

0 1 0 k2

0 0 1 k3

3.) x - 3y -7z = 6

4x + y = 7

2x + 3y +z = 9

4.) 3x - 5z = -1

2x + 7y = 6

x + y + z = 5

... Help T_T this is trial & error probs ... so many, so complicated ... kindly teach me some tips in solving matrices? - Sep 9th 2006, 06:48 AMdan
if the matrix is a 3 x 3 it there is a simple way to solve...

a b c

D= d e f = aei + dhc +gbf - ceg-fha- idb. at least i think that's it^_^

g h i

please note this in only for a 3X3 matrix

dan - Sep 9th 2006, 06:50 AMdan
for a larger matrix you have to cofacter...do a forum serch for "quicklopedia"

i think he talked about that there...

dan - Sep 9th 2006, 04:38 PMThePerfectHacker
No, for a larger matrix you use elementary row operations to row reduce it to echelon form. The complexity of this algorithm is $\displaystyle n^2$, the complexity of using the determinant becomes larger not exactly sure.

What I am trying to say, is that for larger matrices using the determinant is not the most efficient way to do it.

To example, to find the inverse matrix using Gaussian elimination takes $\displaystyle n^3$ steps. To find the inverse using transposes and determinants takes $\displaystyle n!$ ways thus for the first values we have,

1,8,27,64,125...

1,2,6,24,120,...

You can see it is more efficient using Gaussian elimination for matrices of dimension 5. - Sep 9th 2006, 11:59 PMCaptainBlack
- Sep 10th 2006, 12:33 AM^_^Engineer_Adam^_^
ive got an example in our discussion:

2x - y + 3z = 9

3x + y + +2z = 11

x - y + z = 2

A

( Add 2nd row to the 1st row)

2 -1 3 9

3 1 2 11

1 -1 1 2

B

(5 0 5 20) 1/5

3 1 2 11

1 -1 1 2

C

(Add row 2 by row 3)

1 0 1 4

3 1 2 11

1 -1 1 2

D

(Add 2nd row by 1st row)

(1 0 1 4) -2

3 1 2 11

4 0 3 13

E

( Add 3rd row by 1st row)

(1 0 1 4) -3

-1 1 0 3

4 0 3 13

F

(Add 3rd row by 1st row)

1 0 1 4

1 1 0 3

(1 0 0 1) -1

G

(Add 2nd row by 3rd row)

0 0 1 3

1 1 0 3

(1 0 0 1)-1

H

0 0 1 3

0 1 0 2

1 0 0 1

I

1 0 0 1 where x = 1

0 1 0 2 where y = 2

0 0 1 3 where z = 3

looks simple but complicated - Sep 10th 2006, 05:32 AM^_^Engineer_Adam^_^
Nevermind Ive already finished answering it :D

- Sep 10th 2006, 07:10 AMThePerfectHacker
Let me do it my way.

As I said, all you need is to reach a triangluar matrix. (A matrix with the bottom numbers zeros below the diagnol).

The augmented matrix is

[2 -1 3 | 9]

[3 1 2 | 11]

[1 -1 1 | 2]

Thus,

[1 -1 1 | 2]

[2 -1 3 | 9]

[3 1 2 | 11]

Thus,

[1 -1 1 |2]

[0 1 1 |5 ]

[0 4 -1| 5 ]

Thus,

[1 -1 1|2]

[0 1 1|5 ]

[0 0 -5| -15]

Thus,

[1 -1 1|2]

[0 1 1|5]

[0 0 1|3]

And stop right here, why? You will see, first note you got a triangular matrix.

The bottom line says,

z=3

The middle line says

y+z=5 thus y+3=5 thus y=2

The top line says,

x-y+z=2 thus x-2+3=2 thus x+1=2 thus x=1.

Finished!

You see how each line leads to a simple equation.

Doing it like this would say half your time.

This is my 24:):)th post!!! - Sep 10th 2006, 09:10 AMCaptainBlack
Stopping when the matrix is upper-triangular and then using back substitution

is fine if we just have the one equation to solve. But if we have a number of

right-hand sides (as we get if we are using Gaussian elimination to numerically

invert a matrix) it may be more efficient to go all the way and reduce the left-

hand side to an identity matrix.

It may be that this exercises is intended to illustrate the technique for

subsequent application in this manner, so if the original poster has been

asker to reduce the left-hand side to the identity matrix we should respect

that.

RonL - Sep 10th 2006, 09:13 AMThePerfectHacker
- Sep 10th 2006, 09:34 AMCaptainBlack
- Sep 10th 2006, 04:06 PMdan
ph, i'de like to learn more about the 5x5 matrix solving with$\displaystyle n^3$ and [tex]n![tex]

WHAT IS RONG WITH THE CODE ON THE SITE?

do you have a link, or what would i search for on google??

dan - Sep 10th 2006, 04:11 PMThePerfectHacker
- Sep 10th 2006, 06:15 PMdan
- Sep 10th 2006, 06:32 PMThePerfectHacker