Proof that if a square matrix has a row of zeroes, the square matrix is non-invertible. I will like to make sure my mathematical reasoning is sound as I am new to proof.
Can someone with mathematical maturity help me confirm that my proof is sound?


*

*$AX = 0$, given any $A$ square matrix with a row of zeroes.

*A row of zeroes means that there is a non-zero vector $X$ as a
solution to $AX = 0$ (means the column without pivot can be of any solution)

*For the sake of argument, assuming that there is $A^{-1}$: 
$A^{-1}AX = A^{-1}0$
$X = 0$, Since  $A^{-1}A$ gives identity matrix and product of any $0$ gives $0$
Since X has a non-zero vector solution(point 2), by proof of contradiction, $A^{-1}$ does not exist. Since $A^{-1}$ does not exist, by definition, matrix $A$ is non-invertible.
 A: Your proof is correct, but it's overly complicated. You tried to prove that $A$ is singular (i.e. $AX=0$ for some nonzero $X$) and use this fact to prove that $A$ is non-invertible (i.e. $A$ has no inverse), but it's actually easier to prove that $A$ is non-invertible directly.
Suppose the $k$-th row of $A$ is zero. Then the $k$-th row of $AB$ is zero for every matrix $B$. Therefore $AB\ne I$ for every matrix $B$, i.e. $A$ has no inverse.
A: More generally $ A $ will have no inverse if the rows (or columns) of A are linearly dependent. However , I assume you are new to linear algebra and linear dependence has not been introduced to you yet.
In that case, let
AX=Y
where X and Y are column matrices.
If AX=Y has a single solution, then for every Y there exists only one X.
We can find X by $X=A^{-1}Y $
You seem to be aware that if A has a zero row (or many zero rows), the equation $AX=0$ has many solutions..
If $AX=0$ has only a single solution, its the trivial one.. namely X is the zero column.
Hope you get the drift. $A^{-1} $ exists whenever the equation AX=Y has a single solution. A row of zeros (or linearly dependent rows) makes the system have multiple solutions and hence no inverse is possible.
If you are introduced to matrices for solving a system of simultaneous equations, a zero row means you have one less equation in your system.. You need three equations to solve for three unknowns. if you are given only two equations, (which is what a zero row means) , you will obviously have multiple solutions. And whenever there are multiple solutions, system is not invertible.
Therefore, your proof is a direct consequence of existence of non-trivial X such that $AX=0$. If that is justified, your proof is correct.
A: Your proof works. 
Assuming the existence of an inverse leads you to
$AX=0 \Leftrightarrow A^{-1} A X = 0$. As you assumed $X$ to be nontrivial this is a contradiction since $A^{-1} A X = X$. 
What you need to know in order for this to work is that there are nontrivial solutions to $AX = 0$. But since your matrix has a row of zeros this is given. 
