Product of $A$ with the adjoint of $A$: why are all nondiagonal elements zero? Let \begin{align*} A = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ 6 & 8 & 2 \end{pmatrix}. \end{align*} We have $\det(A) = 44$. The cofactor matrix corresponding with $A$ is \begin{align*} C = \begin{pmatrix} -4 & 0 & 12 \\ 28 & -22 & 4 \\ -6 & 11 & -4 \end{pmatrix}. \end{align*} Hence the adjugate matrix of $A$ is \begin{align*} C^T = \begin{pmatrix} -4 & 28 & -6 \\ 0 & -22 & 11 \\ 12 & 4 & -4 \end{pmatrix}. \end{align*} Then \begin{align*} A C^T = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ 6 & 8 & 2 \end{pmatrix} \begin{pmatrix} -4 & 28 & -6 \\ 0 & -22 & 11 \\ 12 & 4 & -4 \end{pmatrix} = \begin{pmatrix} 44 & 0 & 0 \\ 0 & 44 & 0 \\ 0 & 0 & 44 \end{pmatrix} = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & \det(A) & 0 \\ 0 & 0 & \det(A) \end{pmatrix}. \end{align*} 
Now I'm trying to understand in an intuitive manner why all the elements $a_{ij}$ for $i \neq j$ are zero. I understand why the diagonal elements are all $44$. For $a_{11}$ this is just Laplace expansion along the first row, i.e. \begin{align*} \sum_{j=1}^3 (-1)^{1+j} a_{1j} \det(M_{1j}) = \sum_{k=1}^3 a_{1k} C_{1k}, \end{align*} with $C_{1k} = (-1)^{1+j} \det(M_{1j})$.
Now, let's say we compute $a_{12}$. Then we multiply row $1$ of $A$ with column $2$ of $C^T$, or equivalently with row $2$ of the cofactor matrix $C$, since it is the transpose. I don't see why this should always equal zero? Why is $ \begin{pmatrix} 1 & 2 & 4 \end{pmatrix} \cdot \begin{pmatrix} 28 & -22 & 4 \end{pmatrix} = 0 $?

Because this is the Laplace expansion for the determinant of a matrix with two identical rows.
