$AC^T = \det(A)I$ Let A = $\begin{bmatrix}    a_{11} & a_{12}  & \dots & a_{1n} \\ a_{21} & a_{22}  & \dots & a_{2n}  \\ \vdots & \vdots  & \ddots & \vdots \\ a_{n1} & a_{n2} &
\dots & a_{nn} \end{bmatrix}$ 
and the matrix of cofactors of $A$ is 
$$C=\begin{bmatrix}    C_{11} & C_{12}  & \dots & C_{1n} \\ C_{21} & C_{22}  & \dots & C_{2n}  \\ \vdots & \vdots  & \ddots & \vdots \\ C_{n1} & C_{n2} &
\dots & C_{nn} \end{bmatrix}.
$$
I try to understand why $AC^T = \det(A)I$ necessarily. 
Why is it that $a_{11}C_{21} + a_{12}C_{22} + \dots + a_{1n}C_{2n} = 0$?

The expression
$$
\color{red}{a_{11}}C_{21} + \color{red}{a_{12}}C_{22} + \dots + \color{red}{a_{1n}}C_{2n}
$$
is the Laplace expansion along the second row of
$$
\begin{vmatrix}
a_{11} & a_{12}  & \dots & a_{1n} \\ \color{red}{a_{11}} & \color{red}{a_{12}}  & \color{red}\dots & \color{red}{a_{1n}} \\ a_{31} & a_{32} & \dots & a_{3n} \\ \vdots & \vdots  & \ddots & \vdots \\ a_{n1} & a_{n2} &
\dots & a_{nn}
\end{vmatrix}=\color{red}{a_{11}}C_{21} + \color{red}{a_{12}}C_{22} + \dots + \color{red}{a_{1n}}C_{2n}=0.
$$
Edit: take the matrix $A$ and do the determinant expansion along the second row
$$
\begin{vmatrix}
a_{11} & a_{12}  & \dots & a_{1n} \\ \color{blue}{a_{21}} & \color{blue}{a_{22}}  & \color{blue}\dots & \color{blue}{a_{2n}}  \\ a_{31} & a_{32} & \dots & a_{3n} \\ \vdots & \vdots  & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn}
\end{vmatrix}=\color{blue}{a_{21}}C_{21} + \color{blue}{a_{22}}C_{22} + \dots + \color{blue}{a_{2n}}C_{2n}.
$$
Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.
