# Proof Regarding Determinants of a Matrix

Prove the following statement:

If $A$ is an $n$ by $n$ matrix, such that $\sum_{j = 1}^n a_{ij} = 0$, for all $1 ≤ i ≤ n$, then $\det A = 0$ too.

(Sorry I don't know how to format this equation)

Firstly, I've tried to understand what this summation notation means. I think it means that in any $i$-th row, if the coefficients sum up to zero, then the determinant of that matrix will be $0$. I don't know if this is right, but if it is please let me know.

Also, how would I go about proving this statement. It's fairly confusing for me. Thanks.

Hint: if this row-sum property holds (yes, you are interpreting this correctly), consider the vector $x = (1,1,1,\dots,1)^T$. What is $Ax$?
Some Hint: If you consider the Matrix $A$ as $n$ column vectors and one of those column vecors is linearly dependent on the other $n-1$ column vectors, then $\det A$ is zero.
Let $$C_j$$ be the column vectors of $$A$$. This property means $$C_1+C_2+\dots+C_n=0$$. Now the determinant is a multilinear function of the column vectors, i.e. : \begin{align*} \det (C_1+C_2+\dots+C_n, C_2,\dots, C_n)=\det (C_1, C_2,\dots, C_n)+\sum_{j=2}^n\det (C_j, C_2,\dots, C_n) \end{align*} The determinant is also an antisymmetric function of the column vectors, which implies that if any two columns are equal, the determinant is $$0$$. Thus $$\det (C_1, C_2,\dots, C_n)=\det (C_1+C_2+\dots+C_n, C_2,\dots, C_n)=\det (0, C_2,\dots, C_n)=0$$ by multilinearity.