An $n\times n$ matrix $A$ is said to be singular if there exists a nonzero vector $\mathbf{v}$ such that $A\mathbf{v}=\mathbf{0}$.

Otherwise, we say that $A$ is a nonsingular matrix.

Proof.

Let $\mathbf{v}=\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix}$ be the $n$-dimensional vector all of whose entries are $1$.
Then we compute $A\mathbf{v}$.

Note that $A\mathbf{v}$ is an $n$-dimensional vector, and the $i$-th entry of $A\mathbf{v}$ is the sum of the $i$-th row of the matrix $A$ for $i=1, \dots, n$.
It follows from the assumption that the sum of elements in each row of $A$ is $0$ that we have
\[A\mathbf{v}=\mathbf{0}.\]

As $\mathbf{v}$ is a nonzero vector, this equality implies that $A$ is a singular matrix.

Example

To illustrate the proof, let us consider a concrete example.
Let $A=\begin{bmatrix}
1 & -4 & 3 \\
2 & 5 &-7 \\
3 & 0 & -3
\end{bmatrix}$.
Then the sum of each row of $A$ is zero.

Now we compute
\begin{align*}
\begin{bmatrix}
1 & -4 & 3 \\
2 & 5 &-7 \\
3 & 0 & -3
\end{bmatrix}\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
=\begin{bmatrix}
1+(-4)+3\\
2+5+(-7) \\
3+0+(-3)
\end{bmatrix}
=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.
\end{align*}
This shows that the matrix $A$ is singular because we have $A\mathbf{v}=\mathbf{0}$ for a nonzero vector $\mathbf{v}$.

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