If $A_{n \times n}x=b$ has no solutions then $Ax=0$ has infinitely many solutions

Question

If $A$ is matrix of order $n \times n$ over an infinite field and $Ax=b$ has no solutions then $Ax=0$ has infinitely many solutions. Is that true?

Answer 1

Yes. This is follows from the rank-nullity theorem. If $Ax=b$ has no solution, it means that there is a vector that does not belong to the range of $A$. Hence, the dimension of the range of $A$ (i.e., the rank of $A$) is at most $n-1$. So the dimension of the kernel of $A$ (i.e., the nullity of $A$) is at least $1$. So there are infinitely many $x$ such that $Ax=0$.

Answer 2

If $Ax = b$ has no solutions, then $\det(A) = 0$. This is equivalent to say that $\ker(A)\neq 0$, i.e., that the system $Ax=0$ has infinity solutions (its solutions are exactly the subspace $\ker(A)$).