If $$A~=~\\left[~a_{i~j}~\\right]$$ is a square matrix of order n, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefore, $$\\sum^n_{i~=~1}~a_{i~j}~C_{i~j}~=~\\left|~A~\\right|~and~\\sum^n_{j~=~1}~a_{i~j}~C_{i~j}~=~\\left|~A~\\right|$$ $$Given:~\\left|~A~\\right|~=~a_{11}~C_{11}~+~a_{12}~C_{12}~+~a_{13}~C_{13}~\\left[\ ext{~Expanding~along~}R_1~\\right]$$ Now, $$\\left|~A~\\right|~=~a_{12}~C_{12}~+~a_{22}~C_{22}~+~a_{32}~C_{32}~\\left[\ ext{~Expanding~along~}R_2~\\right]~\\left[~a_{12}~,~a_{22}\ ext{~and~}a_{32}\ ext{~are~elements~of~}C_2~\\right]~$$