How to prove that a square matrix's determinant is zero given its row and column properties?

Question

How can I prove the following?

If $A$ is an $n \times n$ matrix such that
$$ \sum\limits_{j=1}^n a_{ij} = 0 $$ for $1 \leq i \leq n$ then $\det A = 0$.

Answer 1

If $\sum_{j=1}^n a_{ij}=0$ for all $1\leq i\leq n$, then in particular $$A\cdot \begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}=\begin{bmatrix}a_{11}\cdot 1 + \cdots + a_{1n}\cdot 1\\ a_{21}\cdot 1 + \cdots + a_{2n}\cdot 1\\ \vdots \\ a_{n1}\cdot 1 + \cdots + a_{nn}\cdot 1 \end{bmatrix}=\begin{bmatrix}0\\0\\\vdots \\ 0\end{bmatrix}$$

So, since the system $A\overrightarrow{x}=\overrightarrow{0}$ does not have a unique solution $\overrightarrow{x}=\overrightarrow{0}$, we can conclude that $A$ is not invertible and thus $\det(A)=0$.

Answer 2

Hint: Compare that with a linear combination of the column vectors $a_i$. What does it tell you about the set of $a_i$ and their parallelepiped?

Answer:

The expression is a linear combination of the column vectors of $A$ with non-zero coefficients. Thus the column vectors are linear dependent and the parallelepiped they span has zero volume, thus $\DeclareMathOperator{det}{det}0 = V=\det(A)$.