# Existance of a matrix that satisfies a condition

1. Jan 20, 2009

### fluidistic

1. The problem statement, all variables and given/known data
A=(2 3, 1 2, 2 5) where the coma separates the rows of the matrix.
Does there exist a matric C such that AC=I? Where I is the 3x3 identity matrix.
2. The attempt at a solution
No. First I note that if it exists then C is a 2x3 matrix. I also note that if AC=I, then C is the inverse of A. But as A is not a square matrix, it doesn't have an inverse, so C cannot exist.
Am I right?

2. Jan 20, 2009

### MathematicalPhysicist

Not exactly.
2 3
1 2
2 5
C should be 2x3, i.e of the form:
a b c
d e f

so you need to solve the system of equations:
2a+3d=1
2b+3e=0
2c+3f=0
2a+3d=0
b+2c=1
c+2f=0
2a+5d=0
2b+5e=0
2c+5f=1

I don't know if the answer is affirmative or not, you need to solve these equations, in general it's possible that even a non square matrix will have a right reciprocal or left one, and both of them need not be the same necessarily.

3. Jan 20, 2009

### Dick

You could also note that your product looks like column(v1,v2,v3)*row(w1,w2,w3) where the v's (rows) and w's (columns) are 2-vectors. You know one of the v's can be written as a linear combination of the other two, say v3=a*v1+b*v2. You want v3.w3=1. What do other matrix entries in the product AC tell you about v1.w3 and v2.w3?

Last edited: Jan 20, 2009
4. Jan 20, 2009

### fluidistic

Ok I think I know where I made an error, it was there :" I also note that if AC=I, then C is the inverse of A."
I'm not really used to linear combinations so I don't get all what you said Dick, but very soon I'll be in it.
Doing the method of loop quantum gravity, I realized from the beggining that the system has no solution.