Prove that the summation of product of co-factor of an element of a matrix with any other element is 0.

Question

Let A = ($a_{ij}$) be an $n×n$ matrix and $A_{ij}$ be the co-factor of $a_{ij}$.
Show that:
$\sum$ $a_{ik}$$A_{jk}$ = det($A$) if i=j, else 0

I couldn't think of any way to do this other than trying to open the co-factor, but I think that would be vigorous.

EDIT- I've tried expanding, but there were a lot of terms, I got confused how I could accommodate them all. Also, as i is not equal to j, there are lots of cases. How can we generalize all that?

Answer 1

Hint

• Replace the column $j$ of the matrix $A$ with column $i$.
• Then use cofactor expansion with regard to column $j$.
• If $j\neq i$, then obviously the result is equal to zero as we're expanding the determinant of a matrix having two columns equal.
• And if $j = i$, this is the classical formula of the determinant expansion according to column $j$ of the matrix $A$. A proof of this formula is provided at Laplace expansion in wikipedia. Another way to prove it is to prove that the determinant is an $n$-linear alternating form of the column vectors. See properties of the determinant.