TITLE: Show the sum of cofactors $C_{ij}$ for $i \neq j$ of $A$ is $=0$
QUESTION [0 upvotes]: Letting $A = (a_{ij})$ of an $n\times n$ matrix. How would I show the following is true:
$$a_{i1}C_{j1}+a_{i2}C_{j2}+\cdots+a_{in}C_{jn}=0$$
I know that $C_{ij} = (-1)^{i+j}M_{ij}$ Where $M_{ij}$ is a minor matrix. I also know that $$\det(A) = \sum^n_{k=1}a_{ik}C_{ik}$$ for $i = j$. 
Many thanks!

REPLY [0 votes]: Presumably you mean for $i \ne j$ (mentioned in the title but not in the body), as for $i=j$ you get $\det(A)$ (as you mentioned). 
The left side of your expression is the Laplace expansion on row $j$ for the determinant of the matrix obtained by replacing row $j$ by a copy of row $i$.  The determinant of a matrix with two identical rows is of course $0$.
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