In the common case, NO ! To determine exactly a linear transformation in a 3 dimensional linear space you have to know the base and you have to know the transformation (the map) of this base. I see here only 2 vectors: K1 and K2. And I do not know if K1 and K2 are linear independents. If this is the case, then use one more vector K3. Then you could use with succes the inverse matrix [k1 k2 k3]-1.

The problem is definitely solvable. It is just that it has infinite solutions. You have 9 unknowns in the A matrix and only 6 equations given by your second and third equation. To find a specific solution, you can arbitrarily fix any 3 of the 9 unknowns. Hope this helps.

Sorry, I won't have said "arbitrarily". But you can fix a column of A to get a specific solution. Or you can fix the diagonal of A, which I think may be much safer.

Peeyush,

if you provide me with a correctly written version the 6th line of your question,

I will provide you with the solution.

Check the attached m file for an example.

Write your problem

A*K1 = K2

A*A*K1 = K3

or more suggestiv, and avoiding the quadratic appearence of A (so that

we are in the realm of linear algebra, despite the fact that non-linear theory

is modern these days)

A*K1 = K2

A*K2 = K3

or even more suggestiv

A*[K1,K2] = [K2,K3] (*)

where [K1,K2] and [K2,K3] are 3 by 2 matrices written down in matlab notation.

Now we have a completely transparent matrix equation:

unknown (3,3)-matrix * known (3,2)-matrix = known (3,2)-matrix.

If the known matrices were of size (3,3) we would get A uniquely by matrix inversion from (*):

A = [K2,K3] * [K1,K2]^-1 (**)

In our case the solution is not unique. As is clear from counting unknowns the general

solution depends on three parameters which are allowed to take any value and nevertheless

A is a perfect solution. So far, so traditional. Now there is a high-light you may not yet have accountered so far:

All linear underdetermined equation systems have just one 'best' solution which is given

(uniquely, as already said) by replacing matrix inversion by 'pseudo inversion', also called

'Penrose inversion'. According to which logic this 'Penrose solution' is the best, this you should look up

in books or the web. This says that you should (*) take as your solution where ^-1 is taken to mean

the pseudo inverse (PseudoInverse[...] in Mathematica, matlab certainly has it too). If you should not only the

natural solution but all solutions you have to activate a linear equation solver for (*).

Ulrich, I like your formulation in (*). I believe the 'Penrose solution' has the least norm. And the MATLAB code for solving A is actually

A=( [ K1' ; K2' ] \ [ K2' ; K3' ] )'

My C++ library, called C+-, allows writing this solution a bit more transparently:

A = (K2&&K3) / (K1 && K2)

i.e. without the many transpositions. Info on C+- :

http://www.ulrichmutze.de/softwaredescriptions/cpmProject.pdf

It is very interesting of what you did. I'll definitely use it as a resource. Thanks!