Proving that if $\mathbf{A}\cdot\mathbf{1} = \mathbf{0}$ then $\mathbf{A}$ has a zero determinant $\mathbf{A}$  is a square matrix of order $n\times n$, $\mathbf{1}$  is the column vector that has all elements are $1$, $\mathbf{0}$  is the column vector that has all elements are zero.
I encounter this problem in a book of matrix algebra and haven't been able to find the solution to it yet. Could anyone please help me on this?
Sorry this is my first post on the site (and also my first post asking for solutions relating to mathematics so there might be many issues in wording and format of notations).
Thanks all for your help!
 A: If $A$ is non-singular, then the system $Ax=0$ only has the trivial solution. 
In this case, you have found a non-trivial solution, hence it must be singular.
A: You can calculate the determinat directly. Let $c_1 , \ldots , c_n$ be the columns of $A$:
$$detA = det(c_1\; c_2 \; \ldots \; c_n) = det(c_1+c_2+\ldots + c_n \; c_2\; \cdots \; c_n)  \stackrel{A\mathbf{1}=\mathbf{0}}{=} det(\mathbf{0} \; c_2 \;\ldots c_n) = 0 $$ 
A: $$\mathbf{A}\cdot\mathbf{1} = \mathbf{0}$$
The terms of  the column vector,  $$\mathbf{A}\cdot\mathbf{1}$$ are simply row sums of matrix $ \mathbf{A}$
$$ a_{11} +a_{12} +... a_{1n}=0$$
$$ a_{21} +a_{22} +... a_{2n}=0\\.\\.\\.$$
$$ a_{n1} +a_{n2} +... a_{nn}=0$$
Thus the columns of $ \mathbf{A}$ are linearly dependent.
Which results  in $$ \det  \mathbf{A}=0$$ 
A: Just in case you happen to be allowed to use eigenvalues: $$A\,\boldsymbol{1}_n =\boldsymbol{0}_n =0\cdot\boldsymbol{1}_n \implies \text{$0$ is an eigenvalue of $A$} \implies \det(A)=0.$$
A: A matrix $A$ having a non-zero determinant is equivalent to a great many things. Among a few are:


*

*The system $Ax = b$ has a unique solution for all $b$.

*The system $Ax = 0$ has only one solution: $x = 0$.

*The scalar $0$ is not an eigenvalue of $A$.


Take your pick! They all imply the above.
A: $$\mathbf{A}\cdot\mathbf{1} = \mathbf{0}$$ 
We also have  $$\mathbf{A}\cdot\mathbf{0} = \mathbf{0}$$ 
That is the system in   $$\mathbf{A}\cdot\mathbf{X} = \mathbf{0}$$ has more than one solution.
This happens only if $\det \mathbf{A}=0$ 
A: Another direct proof (besides trancelocation's): multiply both sides of $0=A\mathbf1$ by $\operatorname{adj}(A)$. Then $0=\operatorname{adj}(A)A\mathbf1=\det(A)I\mathbf1=\det(A)\mathbf1$. Comparing both sides entrywise, we get $\det(A)=0$.
