# How to find unique solution of X

1. Dec 31, 2013

### sudipta4

Hi All,

I have a simple problem

consider two matrix equations
A1X=B1
A2X=B2

A1 and A2 are 3X3 matrix. Moreover, A2 is a positive, definite, symmetric matrix. B1 and B2 are two 3X1 column vectors. The A1, A2, B1, B2 are known. I need to find unique solution of X vector from these two equations. Is there any suitable way to do that?

Can I do it by using Lagrange multiplier? It is not a homework problem. I am struggling this by two days but didn't get success.

Best
Sudipta

2. Dec 31, 2013

### phyzguy

Your system is over-defined, Each of your two matrix equations represents three equations, so you have a total of 6 equations and only three unknowns. Let's assume A1 and A2 are non-singular. Unless:
$$A2^{-1}B2 = A1^{-1} B1$$
there is no solution. If this is satisfied, then
$$X = A2^{-1}B2 = A1^{-1} B1$$

3. Dec 31, 2013

### sudipta4

Hi,
Thank you very much for your reply.

Actually, I want to use first equation is a constraint for second one. If so how do I combine the first equation with the second equation by some constrained minimization method like Lagrange multiplier method.

in fact, my two matrices are
A1=[[-1 -1 0] [1 0 -1] [0 1 1]]
A2=[[a11 a12 a13] [a12 a22 a23] [a13 a23 a33]]

can you give some clue, how do I proceed?

4. Dec 31, 2013

### phyzguy

Do you know B1 and B2? If so, what are they?

5. Dec 31, 2013

### AlephZero

it A2 is positive definite, it s nonsingular, you have a unique solution of A2X = B2.

If that solution does not satisfy A1X = B1, you need to reconsider what you are really trying to do here. You can't "do something impossible" by trying to use A1X = B1 as constraint equations, or whatever.

6. Dec 31, 2013

### sudipta4

I know the values of B1 and B2 but they are different
B1=[-12.212 0.56 0.65]

B2=[-68.49 -68.49 0.0]
Also the A2 matrix is
A2=[[151.8 2.4 -149.4][2.4 151.8 149.4] [-149.4 149.4 298.8]]

These values are related with my system

Last edited: Dec 31, 2013
7. Dec 31, 2013

### phyzguy

It's easy to see that no solutions to your problem exist. Expand your equation A1 X = B1, to get:
-x1 - x2 = b1
x1 - x3 = b2
x2 + x3 = b3

Now add all three equations together, giving 0 = b1 + b2 + b3. Your B1 does not satisfy this equation. Therefore no solutions exist.

8. Jan 1, 2014

### sudipta4

Yes, phyzguy, you are right. I know that. Moreover, A1 is rank deficient and determinant of A2 is 0. therefore, it will not be giving a solution in a simple way. but i need that solution. please help me out in this regard. please give me some clue if anyone aware about this kind of problem.