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Algorithms :: Maths :: Inverse of a Matrix?
By: Raydot U.S.A.  Date: 27/05/2003 00:00:00  English French  Points: 300 Status: Answered
Quality : Excellent
I was asked this question recently and it made my head spin.

Invert the following matrix (I'm doing my best, typographically):

1 - sqrt 3
- ---------
2 2

sqrt 3 1
------- -
2 2

(That's four terms in a 2x2 grid.)

First I had to check to see if it was linear before inverting, HUH? Please be specific, I'd like to learn from this.
By: VGR Date: 27/05/2003 07:53:00 English  Type : Comment
(In fact, your matrix has a dominant diagonal so there are other solutions than this : )

Let's call it "A"

B is the inverse of A if A.B=I (identity matrix)

Let's say that B = ((x,y),(z,t))

thus you come up with a linear set of equations :
x/2 -Sqrt(3).z/2=1
y/2-Sqrt(3).t/2=0
etc

and thus you obtain that B=Sym(A) its symetric matrix
1/2 sqrt(3)/2
-Sqrt(3)/2 1/2
By: Raydot Date: 28/05/2003 02:56:00 English  Type : Comment
Uh..."dominant diagonal?"
By: VGR Date: 28/05/2003 03:11:00 English  Type : Comment
<A HREF="http://asi.insa-rouen.fr/enseignement/siteUV/ananum/05syslindirect.pdf">http://asi.insa-rouen.fr/enseignement/siteUV/ananum/05syslindirect.pdf</a>
By: VGR Date: 28/05/2003 03:13:00 English  Type : Comment
whatever i=1..n, Abs(Aii)>sigma(j=1..n, j<>i) of Abs(Aij)
By: VGR Date: 28/05/2003 03:14:00 English  Type : Comment
this implies that Gauss works with no column change (pivot) and that the matrix is non-singular
It's also one of the preconditions for algorithms named Choleski, Rouge/Noir (Red/Black?) and LDL'
By: Raydot Date: 28/05/2003 04:12:00 English  Type : Comment
Hm...well I don't speak French, although I guess I followed the diagrams. Let me think through this and I'll get back to you. I have a feeling I'm going to have a lot more questions.

What does it mean "column change" and "the matrix is non-singular?"

I'm not dumb, I swear. Math just gets me every time.
By: VGR Date: 28/05/2003 04:32:00 French English  Type : Assist
matrice singulière (I'll translate the best I can)

Une matrice est singulière si on peut trouver une combinaison linéaire des lignes
ou des colonnes qui donne le vecteur 0.

A matrix is singular if you can find a linear combination of the lines or columns which produces the zero vector

For instance, direct methods of resolution of linear equations like Householder's (aka QR) can't work on a singular matrix.

#include <math.h>
#include <conio.h>
#define NMAX 5

void al_prod_mat(double a[NMAX][NMAX],double b[NMAX][NMAX],double c[NMAX][NMAX],int n)

int sl_decomp_qr(double a[NMAX][NMAX],double q[NMAX][NMAX],int n)

void sl_resol_qr(double q[NMAX][NMAX],double r[NMAX][NMAX],double b[NMAX],double x[NMAX],int n)

main()
{
double a[NMAX][NMAX],q[NMAX][NMAX];
double b[NMAX],x[NMAX];
int i,j,n;
int err;
n=4;
clrscr();
printf("Méthode de la Décomposition QR\n");
a[1][1]=8;a[1][2]=-4;a[1][3]=3;a[1][4]=7;b[1]=12;
a[2][1]=4;a[2][2]=2;a[2][3]=-6;a[2][4]=4;b[2]=1;
a[3][1]=-16;a[3][2]=6;a[3][3]=-2;a[3][4]=-15;b[3]=-19;
a[4][1]=6;a[4][2]=10;a[4][3]=-15;a[4][4]=10;b[4]=1;
printf(" A(1)\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%14.8e ",a[j]);
printf("\n");
}
err=sl_decomp_qr(a,q,n);
if (err==1)
{
printf(" Q\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%14.8e ",q[j]);
printf("\n");
}
sl_resol_qr(q,a,b,x,n);
printf("Solution:\n");
for(j=1;j<=n;j++)
printf("x%d = %22.16e\n",j,x[j]);
printf("\n");
}
else printf("Erreur, matrice singulière\n");
}
By: VGR Date: 28/05/2003 04:33:00 English  Type : Comment
Jordan's method (evolving on Gauss's) also can't work on a singular matrix, so determining that a matrix having a dominant diagonal is non-singular is essential and very useful as the criterion is very simple indeed.

#

#include <math.h>
#include <conio.h>
#define NMAX 5
#define N2MAX 6

int sl_jordan_aff(double a[NMAX][NMAX],double b[NMAX],int n)

main()
{
double a[NMAX][NMAX],b[NMAX];
int i,j,n,err;
clrscr();
n=4;
printf("Méthode de Jordan\n");
a[1][1]=8;a[1][2]=-4;a[1][3]=3;a[1][4]=7;b[1]=12;
a[2][1]=4;a[2][2]=2;a[2][3]=-6;a[2][4]=4;b[2]=1;
a[3][1]=-16;a[3][2]=6;a[3][3]=-2;a[3][4]=-15;b[3]=-19;
a[4][1]=6;a[4][2]=10;a[4][3]=-15;a[4][4]=10;b[4]=1;
printf(" A(1) b(1)\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%12.6e ",a[j]);
printf("I %12.6e \n",b);
}
err=sl_jordan_aff(a,b,n);
if(err==1)
{
printf("Solution:\n");
for(i=1;i<=n;i++)
printf("x%d =%23.16e \n",i,b);
}
else printf("Erreur, matrice singulière");
}
By: VGR Date: 28/05/2003 04:36:00 English  Type : Comment
and for Gauss' story about "no column swap required", it's the fact that Gauss 's method swaps two lines when a null pivot is encountered. A non-singular matrix won't have a null pivot, so the method is fast and efficient, and guaranteed to work.

regards

PS : It's been a looong time since I did some linear algebra ;-)
By: GwynforWeb Date: 28/05/2003 06:41:00 English  Type : Answer
What on earth is every body talking about? this is a perfectly well behaved matrix with a unique inverse. This is done simple row opertations, I assume you have done some linear algebra before but have forgotten. This the standard way of inverting a matrix You do it by setting up the following


1/2 -sqrt(3)/2 | 1 0
|
sqrt(3)/2 1/2 | 0 1


ie the original matrix and the identity, now subract suitable amounts of rows from each other on the left hand ride to get the identity, do the same to the right hand side. The RHS will end be the inverse

subtract sqrt(3) times the 1st row from the second

1/2 -sqrt(3)/2 | 1 0
|
0 2 | -sqrt(3) 1


add sqrt(3)/4 times the 2nd row the 1st


1/2 0 | 1/4 sqrt(3)/4
|
0 2 | -sqrt(3)/2 1


multiply both row 1 by 2 and row 2 by 1/2

1 0 | 1/2 sqrt(3)/2
|
0 1 | -sqrt(3)/2 1/2



the LHS is now the identity and the RHS matrix ie


1/2 sqrt(3)/2

-sqrt(3)/2 1


is the inverse
By: VGR Date: 28/05/2003 07:34:00 English  Type : Comment
sorry but it's wrong.

It's 1/2 at the bottom right

you must have forgotten things too ;-)
By: GwynforWeb Date: 28/05/2003 08:11:00 English  Type : Comment
the last lines should read (the inverse is correctly given in the lines above)

1/2 sqrt(3)/2

-sqrt(3)/2 1/2
By: GwynforWeb Date: 28/05/2003 08:15:00 English  Type : Comment
VGR , "but have forgotten" was not addressed to you.
By: VGR Date: 28/05/2003 08:34:00 English  Type : Comment
ho ? too bad :D

!!!FIGHT!!!

OK, the story so far is that if you generalize your question to more than a 2x2 matrix, you'll have a lot of work in applying manual techniques presented here. Inversing a matrix and solving the equatiob A.B=I is the same, the latter leading to a set of classical linear equations.

I think it's good to know "how to recognize" special matrices because it can save a lot of work, for example saying "it's singular, so it's not inversible" is a good answer if you can prove it :D and it took you 30 seconds ;-)
By: GwynforWeb Date: 28/05/2003 11:23:00 English  Type : Comment
VGR

A matrix is singular if its determinant is zero so for a 2x2

a b

c d

the determinant is ad-bc. It is easily calculated for for any nxn matrix


:-)
By: VGR Date: 28/05/2003 16:52:00 English  Type : Comment
yes, that's Cramer's method, but it's expensive (computing multiplications, adding them)

that's EXACTLY why the theorem of the dominant diagonal is wonderful : with simple ABS() additions, you reach the same result :D :D :D

whatever i=1..n, Abs(Aii)>sigma(j=1..n, j<>i) of Abs(Aij)
By: GwynforWeb Date: 29/05/2003 23:49:00 English  Type : Comment
VGR,
Diagonallly dominant => non-singular but non-singular !=> Diagonallly dominant

eg 1 5

5 1

Diagonal dominance is only of any real practical use for sparse and patterned matrices. During the inversion process it becomes apparent if the matrix is singular when a row zeroes occurs.

All I think Raydot wanted was to kmow how to invert a 2x2 matrix.

:-)
By: VGR Date: 30/05/2003 00:28:00 English  Type : Comment
non-singular !=> Diagonallly dominant

I never wrote this 8-((

"During the inversion process it becomes apparent if the matrix is singular when a row zeroes occurs."

I totally disagree. You did work for nothing. It's better to know beforehand if inversion/resolution is possible.
By: Raydot Date: 30/05/2003 01:23:00 English  Type : Comment
Well, you're both right. Not about the how, but the why. I'd like to learn how to invert *any* matrix, and since the goal is to understand it as a programmer, computational expense is significant. But Gwyn's answers are a wee bit earier to follow.

Could someone explain what a "dominant" is? Is that te row of ones running diagonally?
By: GwynforWeb Date: 30/05/2003 01:31:00 English  Type : Comment
It takes as many computations to evaluate the determimant as it does to as it does to invert a matrix, unless you are very lucky and and can apply tests such as diagonal dominance.

All the alogrithms am I aware of proceed with the inversion and halt if a singularity is detected. ie inversion and detection of a singularity are performed simultaneously.

Diagonal dominance is a very useful test in for sparse patterned matices that frequently occur in the the numerical solution of differential equations, but it can not be applied to any matrix because non-singular !=> diagonallly dominant

:-)


Raydot,
VGR and I have gone well beyond your original question (as we usually do). If you to want invert a matrix, use the method I described. If it is not possible through row opertions to get the identity matrix on the LHS then the matrix does not have an inverse, not all matrices have inverses.

:-)
By: GwynforWeb Date: 30/05/2003 01:44:00 English  Type : Comment
Raydot,
I have just seen your last post. If your are learning how to invert a matrix then the method I have described is just about the simplest there is and is easily programmed. I would image there is masses of code on the web for it.

Diagonal dominance is not some thing you want to bother with just yet. For the record though a matrix is called strictly diagonally dominant if for each row the absolute value of the diagonal element is greater than the sum of absolute values of all the other values in the row. If this is the case then the matrix will have an inverse, but just because it is not diagonally dominant does not mean that it does not have an ivnverse.

eg Diagonally dominant matrix

-3 2 0

-1 4 2

3 -2 6

because

3 > 2

4 > 1 + 2

6 > 2 + 3

but again as I said do not worry about this idea yet, you need to learn how to invert a matrix first.



By: GwynforWeb Date: 30/05/2003 02:03:00 English  Type : Comment
Raydot,
Try inverting the folllowing, the numbers work out nicely, (one of them does not have an inverse)

(a) 1 0
1 1

(b) 1 1
1 2

(c) 1 3
2 7

(d) 2 6
1 4

(e) 2 6
4 12
By: VGR Date: 30/05/2003 02:14:00 English  Type : Comment
@gwyn : because you think that computing the (Cramer's) determinant is necessary to demonstrate a non-singularity of the matrix (it's true), while I demonstrated since my second post that the matrix being diagonally dominant is enough. Computations pass from 0(n²) in additions of multiplications to O(n²) in additions of Absolute values. It's ***a lot*** different.

@raydot for "what is dominant" : Comment from VGR Date: 05/28/2003 10:13AM PDT
By: VGR Date: 30/05/2003 02:19:00 English  Type : Comment
@gwyn : the algorithm you described is (unless I'm wrong) Gauss's ;-)
Of course it's easily programmable. It's also the worse of all algorithms :D
And deciding which method to program for a given class of matrices (like the ones for linear differential equations) is ***also*** done by mathematically analysing the matrices (eg proving that they are diagonally dominant) ***on paper*** and then only to program the ***optimal algorithm***

This is ***always*** better than running a "stupid" gaussian resolution which will stop after the 1022434th iteration saying "sorry, I can't continue, this is a non-inversible [singular] matrix"
By: GwynforWeb Date: 30/05/2003 02:34:00 English  Type : Comment
VGR,
I do not know what you are crying about! I know that strict diagonal dominance is sufficient for a non singularity but it is not necessary. Clearly if you have strict diagonal dominace then you can proceed without worrying about a singularity with an algorithm that will detect a zero determinant anyway (all inversion algorithms detect a singularity).
Because a matrix it is not diagonally dominant does not tell you anyhing and you have wasted your time before proceeding with an algorithm that will detect the singularity anyway!

In general trying to detect a singularity before inversion is a waste of time, the inversion algorithm will find it and terminate for you.
By: GwynforWeb Date: 30/05/2003 02:51:00 English  Type : Comment
VGR,
Just sen your last post. Gauss's method is the simplest algorithm and was designed ( and is the best) for paper and pencil calulations in that it is easy to keep track of errors. There are a multitude of standard techniques for inverting matrices, most of which on the computer are faster than Gauss in it's pure form. When dealing with with a few small inversions it does not matter what you use. When inverting large matrices of a particular class (eg those from a differential equation problem) it is common practice to design a routine for that set of matrices. Gauss is commonly used as a building block in this process.

Diagonal dominance is primarily of theoretical interest and can often reassure the worker that matrix formulation of the problem is correct. It is of little computational importance.

PS the use of *'s is ***** rude and uncharacteristic of our previous discussions
By: VGR Date: 30/05/2003 02:53:00 English  Type : Comment
I don't agree, but I'm only specialized in numerical analysis and computing
By: VGR Date: 30/05/2003 02:56:00 English  Type : Comment
regarding the use of *'s : for us here, it means just a replacement for bold, underligned, italic and other emphasis tags, nothing agressive.
By: GwynforWeb Date: 30/05/2003 03:21:00 English  Type : Comment

"I don't agree, but I'm only specialized in numerical analysis and computing"

so am I. Give me an example of a numerical algortim where an impotant part of it is to check for diagonal dominance before proceeding. The only one I think of is possibly the relaxation method for equations systems, but this is usually done during the decision making process for which algorithm to choose anyway.
By: Raydot Date: 30/05/2003 03:32:00 English  Type : Comment
You guys are awesome, I love it! I'll take a stab at that exercise a bit later and get back to you. Again, thanks for the help.
By: VGR Date: 30/05/2003 03:36:00 English  Type : Comment
The "quoting answer" ;-)

1) "this is usually done during the decision making process for which algorithm to choose"

2) "deciding which method to program [...] and then only to program the optimal algorithm"

3) "I do not know what you are crying about! "
By: GwynforWeb Date: 30/05/2003 04:12:00 English  Type : Comment
??? You are making no sense, the relaxation method for linear systems is the only algorthim I can think of where diagonal dominance is usually checked manually before even coding it up, and that algorthim is not for matrix inversion any way.
By: VGR Date: 30/05/2003 04:19:00 English  Type : Comment
1 and 2 => we agree
3 => réponse du berger à la bergère
By: GwynforWeb Date: 30/05/2003 04:46:00 English  Type : Comment
?
By: _TAD_ Date: 30/05/2003 05:00:00 English  Type : Comment


okay, bear with me here since it's been at least 6 years since my vector equations class (and I haven't used it since). If you write a program that does vector dot and vector cross products, I think a simple solution is to multiply your matrix by the inverse identity matrix.

the matrix:

(3x3)

10, 5, 3
7, 3, 9
1, 3, 5

CROSS

(3x3)
1, 0, 0
0, 1, 0
0, 0, 1

should give you:

(3x3)

10, 5, 3
7, 3, 9
1, 3, 5



BUT... Doesn't this give you the inverse?

(3x3)
10, 5, 3
7, 3, 9
1, 3, 5

CROSS

(3x3)
0, 0, 1
0, 1, 0
1, 0, 0
By: _TAD_ Date: 30/05/2003 05:09:00 English  Type : Comment


Oh boy.. it has been far too long...
By: mrdtn Date: 03/06/2003 13:45:00 English  Type : Comment
No one mentioned that the matrix is the 60 degree rotation matrix applied to a vector in a plane. Perhaps the observation that the inverse would appropriately be the negative 60 degree rotation matrix was more the point -- and that the effect of rotating a vector by 60 and then subsequently by -60 gives you the same thing you started with -- equivalent to multiplying by the identity matrix.

If your intent is to code up a general matrix inversion routine, GwynforWeb's 5/28 1:41PM posting is the best approach. I coded such a routine years ago, and unfortunately don't have the source to provide for you. The approach is simply to diagonalize the LHS, which is pretty straightforward from a coding perspective. If the LHS cannot be diagonalized, the matrix is sigular and has no inverse.
By: Raydot Date: 04/06/2003 02:45:00 English  Type : Comment
Haven't forgotten, I was out of town for the weekend (although, for all I know, I suppose I might have been in your town). Will get back, promise...
By: debiansid Date: 07/06/2003 07:06:00 English  Type : Comment
to simplify the prob, replace sqtr(3)/2 with sin(60), -sqtr(3)/2 with cos(120), 1/2 with sin(120). This should simplify the equations you get.

You can put say, x=60

so you get the matrix as:

cos(x) cos(2x)
sin(x) sin(2x)

now if the inverse is {(a,b),(c,d)} then we get 2 sets of equations as:

a . cos(x) + c . cos(2x) = 1
a . sin(x) + c . sin(2x) = 0

and

b . cos(x) + d . cos(2x) = 0
b . sin(x) + d . sin(2x) = 1

Now with some simple trigonometry, the solution becomes simple and, more importantly, clean.
By: VGR Date: 07/06/2003 07:13:00 English  Type : Comment
mrdtn already said this, and went further than you 8-)

if you turn the matrix into a rot(Pi/3) rotation matrix, then its inverse is Rot(-Pi/3)

you made a mistake in guessing cos(2x) and sin(2x) because you completely forgot the modulo in the resolution of trigonometric equations, and made a mistake :

Halas for you, sin(60°)=sin(120°) and cos(120°)=-1/2 and not 1/2
By: GwynforWeb Date: 07/06/2003 07:21:00 English  Type : Comment
These angle solutions only work for specific types of matrices, I think Raydot wants a general method that is comprehensible.
By: VGR Date: 07/06/2003 07:55:00 English  Type : Comment
one of the god practices in Maths is also to recognize at first glance particular entities and treat them in the most efficient way :D
Generally speaking or programmatically speaking, I agree with you, of course.
By: debiansid Date: 07/06/2003 08:18:00 English  Type : Comment
oops... I think I misread it. Sorry. but I still think (rectifying the sign problem) that the angles approach will make this particular problem much cleaner.
By: debiansid Date: 07/06/2003 08:33:00 English  Type : Comment
Hey wait... I was right!!!

cmon man... sin(90+x)=sin(x). So sin(90+30)=sin(30). And cos(90+x)=-cos(x). So cos(90+30)=-cos(30).

So sin(120)=sin(30) and cos(120)=-cos(30)

So my initial argument holds.
By: VGR Date: 07/06/2003 08:39:00 English  Type : Comment
not at all.

quoting you :
You can put say, x=60

so you get the matrix as:

cos(x) cos(2x)
sin(x) sin(2x)

thus you say that the original matrix is
Sqrt(3)/2 -1/2
1/2 Sqrt(3)/2

whereas it is not. it doesn't contain a "minus 1/2"...
By: VGR Date: 07/06/2003 08:41:00 English  Type : Comment
In fact, you wrote the matrix would be :

1/2 -1/2
Sqrt(3)/2 Sqrt(3)/2

even worse ;-)


And as you wrote, cos(120)=-cos(30) thus "replace -sqtr(3)/2 with cos(120), 1/2 with sin(120). " is false for two reasons I think I don't have to repeat again and again ;-)
By: debiansid Date: 07/06/2003 08:55:00 English  Type : Comment
oh.... simple trigonometry man...

sin(30)=1/2 .............(1)
sin(60)=sqrt(3)/2 .............(2)

cos(60)=1/2 .............(3)
cos(30)=sqrt(3)/2 .............(4)

sin(120)=sin(90+30)=sin(30)=1/2 .............(5)
cos(120)=cos(90+30)=-cos(30)= -sqrt(3)/2 .............(6)

Now we have x=60

so:

cos(x)=1/2 .......... from (3)
cos(2x)= -sqrt(3)/2 .......... from (4,6)

sin(x)=sqrt(3)/2 .......... from (2)
sin(2x)= 1/2 .......... from (1,5)

So my Matrix is seen as:

cos(x) cos(2x)
sin(x) sin(2x)

Substituting I get:

1/2 -sqrt(3)/2
sqrt(3)/2 1/2


Which is, what I believe, we want.

Basic trigonometry. But thanks, you're making me practice for my entrance tests.
By: debiansid Date: 07/06/2003 09:20:00 English  Type : Comment
apologies.... the matrix would be:


cos(x) -sin(2x)
sin(x) -cos(2x)


I think u would agree with me now VGR... :) (It's 4:40 at night man... pardon me)
By: Xxavier Date: 08/06/2003 07:49:00 English  Type : Comment
VGR,
One of the good practices of teaching something new to someone is not to home in special cases, but to try to give the person an understanding of the geneal concept they are dealing with. Secondly if you are going to deal with special cases then do the job properly. Much of the above is at best a very poor treatment of the special propoerties of the matrix :-)

As mrdtn pointed out that the matrix is in fact the 60o rotataion matrix. Ie

cos(60) -sin(60)

sin(60) cos(60)

the inverse of this is again as mrdtn pointed is clearly the the -60o rotation matrix, ie

cos(-60) -sin(-60)

sin(-60) cos(-60)

which is

cos(60) sin(60)

-sin(60) cos(60)


Raydot:- do not let this rotation matrix stuff confuse you
By: GwynforWeb Date: 08/06/2003 08:30:00 English  Type : Comment
oops,
The above is me from a place I am doing some work this week. I am sorry Xxavier I forgot you also had just got an account. I should have checked more carefully who was logged in on the machine.

GfW
By: Xxavier Date: 08/06/2003 12:34:00 English  Type : Comment
GwynforWeb,
You should not have said anything then I could have impressed everybody with my new found knowledge of Mathematics!! ( what is a matrix anyway?)
By: Raydot Date: 09/06/2003 03:14:00 English  Type : Comment
Wow, I turn my back for a second and look what happens! I really don't get one thing that was said about all that trig. I'll just keep reading from 6/4/03 upward, if it's all right with y'all.
By: akshayxx Date: 10/06/2003 05:06:00 English  Type : Comment
>>Basic trigonometry. But thanks, you're making me practice for my entrance tests.
just out of curiosity, which entrance test , what country?
By: debiansid Date: 10/06/2003 05:13:00 English  Type : Comment
India, MCA Entrance (Now don't mock me for the mistakes I made ok)
By: akshayxx Date: 10/06/2003 18:24:00 English  Type : Comment
not mocking .. as u must have made out from my login-id .. me too an indian .. so was just curious , as i believe the word 'entrance-exam' is used more in India than anywhere else in the world.
anyways good luck with ur exams ..
which city though ?
By: akshayxx Date: 10/06/2003 18:30:00 English  Type : Comment
and ya naturally .. with the mention of "the word" .. and "the time" ( 4 am in night,.. shudnt it be morning !!) ...i cudnt help remembering 'my nights'.
By: debiansid Date: 10/06/2003 18:33:00 English  Type : Comment
thanx
I'm from good ol' mumbai!!

Saw your profile by the way. IITian eh!! Cool.

What linux do u use?? (hope its something other than redhat.)
By: debiansid Date: 10/06/2003 18:34:00 English  Type : Comment
as u might have seen from my nick, me a great debian fan.
By: akshayxx Date: 10/06/2003 18:38:00 English  Type : Comment
bad news for u ...from my days ( not too old) .. i am used to redhat ,
anyways, it doesnt matters much.. i anyways customise it to the looks/feels i like :) ..
i guess we are disturbing people in this post .. may be if have anything to talk then we can post it in lounge..
By: Deti Date: 26/07/2003 22:05:00 English  Type : Comment
In my calculation:

Det(A) = 1/4 + 3/4 = 1

Calculating invert Matrix:

D[1,1] = (-1)^(1+1) * 1/2 = 1/2 (identic)
D[1,2] = (-1)^(1+2) * (sqrt(3)) / 2 = (-sqrt(3)) /2 (identic)
D[2,1] = (-1)^(2+1) * -((sqrt(3)) /2) = (sqrt(3)) / 2 (identic)
D[2,2] = (-1)^(2+2) * 1/2 = 1/2 (identic)

Final invert Matrix

We build matrix from top D elements and multiply to 1/det(A). Because Det(A) = 1 so 1/Det(A) = 1 (identic).

Invert Matrix of this is the same Matrix.
By: Raydot Date: 26/07/2003 22:19:00 English  Type : Comment
Yeah, you guys went WAY beyond what I needed. Thank you.

VGR, points for the algorithm -- which I wasn't looking for but which I found useful nonetheless. Gwynn, the rest is yours.
By: GwynforWeb Date: 27/07/2003 04:09:00 English  Type : Comment
Raydot : Thanks for the points. GfW :-)

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