What is the connection between the invertibility of a matrix the kernel of the matrix

If the kernel of a matrix A only has the zero vector (ker(A) = {0})

Then why is the matrix A considered invertible because of that?

I know that if ker(A) = 0, then there is only one unique solution of the matrix( assuming that the matrix is made up of a system of linear equations) but what does this tell me about whether or not the matrix is singular?

Can someone explain the logic behind this?

• If $\det(M) \neq 0$, then there is a unique solution to the relevant system of equations. If $\det(M)=0$, then things get complicated, Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Jan 16, 2018 at 19:58

The word "invertible" and its synonym "non-singular" are reserved for linear systems that have unique solutions. Say that the system $Ax = b$ has a solution. Denote such a solution by $x_0.$ Let $x_h$ (the $h$ stands for "homogenous") denote a nonzero solution to the system $Ax =0$ (called the "homogenous system"). It is true that $x_0$ and $x_0 + x_h$ are two separate solutions to the system $Ax=b.$ So, if there is any nonzero vector in $\ker A,$ then $Ax=b$ has two separate solutions, meaning that it cannot be invertible.
A matrix is invertible if there is a unique solution to the problem $Ax=b$. If there is another vector $x$ besides zero where $Ax=0$ then there is not a unique solution to that problem.
In general, for any function $f:V\to W$ that takes vectors to vectors to have an inverse, for every vector $b\in W$, there must be exactly one element $x\in V$ such that $f(x)=b$. Applying this to the function $f: V\to V$ given by $f: x\mapsto Ax$, this means that for $f$ to have an inverse the system $Ax=b$ must have a unique solution for every $b\in W$. This is precisely the case when the null space of $A$ is trivial.