Kindly Help me plss!!! Matrices

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Sep 9th 2006, 02:04 AM
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 AM
dan
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 AM
dan
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 PM
ThePerfectHacker
Quote:

Originally Posted by dan
for a larger matrix you have to cofacter...do a forum serch for "quicklopedia"
i think he talked about that there...

dan

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 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
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.

The problem looks to me like an exercise in Gaussian elimination.

RonL
• Sep 10th 2006, 12:33 AM
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
• Sep 10th 2006, 07:10 AM
ThePerfectHacker
Quote:

ive got an example in our discussion:

2x - y + 3z = 9
3x + y +2z = 11
x - y + z = 2

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 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
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!!!

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 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
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

That can only happen if that system is not "well-behaved".
• Sep 10th 2006, 09:34 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
That can only happen if that system is not "well-behaved".

Whatever the merits of the algorithm (and no algorithm is perfect) the OP
seems to being asked to do it that way.

RonL
• Sep 10th 2006, 04:06 PM
dan
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 PM
ThePerfectHacker
Quote:

Originally Posted by dan
WHAT IS RONG WITH THE CODE ON THE SITE?

LaTeX is down because of repairs.
• Sep 10th 2006, 06:15 PM
dan
Quote:

Originally Posted by ThePerfectHacker
LaTeX is down because of repairs.

ok thanks,

by the way...i thought it was, "there are only 10 types of people: those who know binary and those who don't".
mayby you use a different version in NY than we use in NE

`dan
• Sep 10th 2006, 06:32 PM
ThePerfectHacker
Quote:

Originally Posted by dan
ok thanks,

by the way...i thought it was, "there are only 10 types of people: those who know binary and those who don't".
mayby you use a different version in NY than we use in NE

Try to stay on topic please, during these past days there was not off topic talk. Let us keep it down.

Just to answer you, that is the third time someone told me that joke is supposed to be ten, yes, I know. But the joke it that it is not.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last