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.