If $N = 1$, the case is clear, so we can assume that $N \geq 2$. Suppose that $a_k = \boldsymbol{0}$ for some $k \in \{ 1, \ldots, N \}$. Using
R5517: Cofactor partition for a complex square matrix, we have
\begin{equation}
\text{Det} A
= \sum_{n = 1}^N A_{k, n} C_{k, n}
= \sum_{n = 1}^N 0 \cdot C_{k, n}
= 0
\end{equation}
If instead $b_k = \boldsymbol{0}$, then the same result implies
\begin{equation}
\text{Det} A
= \sum_{n = 1}^N A_{n, k} C_{n, k}
= \sum_{n = 1}^N 0 \cdot C_{n, k}
= 0
\end{equation}
Since $k \in \{ 1, \ldots, N \}$ was arbitrary, we are done. $\square$
