Proving det(AB)=det(A)det(B)

1. Feb 17, 2013

digipony

1. The problem statement, all variables and given/known data

Prove the relation det(AB)=det(A)det(B)

2. Relevant equations

det A = $\sum_{k}A_{kq} C_{kq}$

3. The attempt at a solution

Here is what I have done:

(detA)(detB)= $\sum_{k}A_{kq} C_{kq}$ $\sum_{k}B_{kq} C_{kq}$
= $\sum_{k}(A_{kq}B_{kq}) C_{kq}$
= $\sum_{k}(AB)_{kq} C_{kq}$
=det(A+B)

My question is: is it mathematically proper to factor out the Ckq in the summation, or am I breaking some matrix/summation rule here?

Thanks!

Last edited: Feb 17, 2013
2. Feb 17, 2013

Ray Vickson

This is riddled with serious errors. First: the cofactors $C_{ij}$ depend on the matrix, so you have one cofactor $C_{ij}(A)$ for matrix $A$ and a different one $C_{ij}(B)$ for matrix $B$. Second, you cannot use the same summation index in both factors, so you need to write something like
$$\det(A) \det(B) = \sum_{k} a_{kp}C_{kp}(A) \sum_{m} b_{mq} C_{mq}(B),$$
and this does not really lead anywhere.

3. Feb 17, 2013

digipony

Ok, I will try to prove it without summation notation. Thank you

4. Feb 17, 2013

Zondrina

Why don't you try considering two cases, one where A is singular and another where A is nonsingular?