Factoring Out Ckq in Summation: Proper or Breaking Rules?

[digipony]

Homework Statement

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

Homework Equations

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

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!

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

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!

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.

 

[digipony]

this does not really lead anywhere.

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

 

[STEMucator]

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