# Hard matrix proof

1. Nov 26, 2009

### thomas49th

1. The problem statement, all variables and given/known data
Prove that the product of two (nxn) non-singular matrixes A and B is non singular and that (AB)^{-1} = B^{-1}A^{-1}

2. Relevant equations

3. The attempt at a solution
Well AI = A^{-1)

we could perhaps try asing AA^{-1} and BB^{-1} ??

Hopeless. What type of things do you look for in a question like this! Just quote rules until you get somewhere?

Thanks
Thomas

2. Nov 26, 2009

### Petek

For the first part, can you use determinants in the proof?

3. Nov 26, 2009

### Hurkyl

Staff Emeritus
That's not unreasonable. A great many problems boil down to little more than citing definitions. If you know what it means to be an inverse, it's almost trivial to check that if B^{-1}A^{-1} is an inverse of (AB).

4. Nov 26, 2009

### Count Iblis

If you want to use determinants then you would have to understand the derivation of the product formula for determinants. Otherwise any proof based on that would be pointless. Given the question in the OP, it is unlikely that he is able to write down the proof of the product formula for determinants from scratch.

So, he should either produce a proof without using determinants, or study determinants first (after which the questions he wanted to answer here would become trivial anyway).

Hint for proof without determinants: try to prove that if AB is singular then either A or B must be singular.

5. Nov 26, 2009

### Office_Shredder

Staff Emeritus
Non-singular is the same as having an inverse. Is B-1A-1 the inverse of AB? If so, you're done

6. Nov 26, 2009

### HallsofIvy

Staff Emeritus
In particular, what is (B-1A-1)(AB)?