Let $A$ be $n\times n$ matrix and there is at least one non trivial solution for the system $Ax=0$.
For any real column vector $b$ with n components,the equation $Ax=b$ has:
$1.$Unique solution.
$2.$Finite many solutions.
$3.$Infinitely many solutions.
$4.$No solution.
I think first and second options are false.
I have problem about the third and fourth options and I think both could be true but I am not sure.
What you will suggest the answer?
Thanks
You're correct: $Ax = b$ will either have infinitely many solutions or no solutions. For example, take $n = 2$ and consider: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ Notice that not all columns of $A$ contain a pivot, so $Ax = 0$ has (infinitely) many (nontrivial) solutions. Yet if: $$ b = \begin{bmatrix} 7 \\ 1 \end{bmatrix} $$ then $Ax = b$ has no solution. On the other hand, if: $$ b = \begin{bmatrix} 7 \\ 0 \end{bmatrix} $$ then $Ax = b$ has infinitely many solutions.