Question

Prove that The sum of the product of the elements of any row (or column) with the corresponding cofactors of elements of any other row (or column) is zero.
i.e. $a_{11}{A}_{21}+a_{12}{A}_{22}+a_{13}{A}_{22}+a_{13}{A}_{23}=0$

Answer 1

$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$

$A_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{32}& a_{33}\end{array}\right|$ From the definition of co-factor

$=(-1)(a_{12}{a}_{33}-a_{32}{a}_{13})=a_{32}{a}_{13}=a_{12}{a}_{33}$

$A_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}{a}_{11}& a_{13}\\ a_{31}& a_{33}\end{array}\right|=a_{11}{a}_{33}-a_{31}{a}_{13}$

$A_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{31}& a_{32}\end{array}\right|=a_{31}{a}_{12}-a_{32}{a}_{11}$

$a_{11}{A}_{21}+a_{12}{A}_{22}+a_{13}{A}_{23}$

$=a_{11}(a_{32}{a}_{13}-a_{12}{a}_{33})+a_{12}(a_{11}{a}_{33}-a_{31}{a}_{13})+a_{13}(a_{31}{a}_{12}-a_{32}{a}_{11})$

$=a_{11}{a}_{32}{a}_{13}-a_{11}{a}_{12}{a}_{33}+a_{12}{a}_{11}{a}_{33}-a_{12}{a}_{31}{a}_{13}+a_{13}{a}_{31}{a}_{12}-a_{13}{a}_{32}{a}_{11}$.