Existence of a unique solution implies that the determinant is not $0$.

Let $$AX=Y$$ be a non-homogeneous linear system where $$A$$ is an $$n\times n$$ matrix. If this system has a unique solution, can we deduce that $$\det A\neq 0$$ directly from cramer's rule?

Using the relation $$x_i\det A = \det A_i$$, if $$\det A = 0$$ then each $$A_i=0$$. Then I have no idea of what to do next.

I know that the existence of a non-trivial solution to the homogeneous system $$AX=0$$ implies that $$\det A = 0$$. But is the result even true in the non-homogeneous case?

It is true. Not sure it is a good idea to use Cramer's rule because if you use it then you assume the determinant is non zero. Anyway, let's say $$u$$ is the only solution to the system $$Ax=y$$. Suppose $$\det(A)=0$$. Then the system $$Ax=0$$ has a non trivial solution, let's call it $$z$$. Now let $$v=u-z$$ which implies $$z=u-v$$. Since $$z\ne 0$$ we know that $$u\ne v$$. But now look what we get:
$$0=Az=A(u-v)=Au-Av$$
So $$Av=Au=y$$. Since $$v\ne u$$ we get that the system $$Ax=y$$ has two different solutions and this is a contradiction.