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In the book of Linear Algebra by Werner Greub, at page 87 section 3.7, it says that

If a corresponding homogeneous system of an inhomogeneous system possesses only the trivial solution, the inhomogeneous system has a solution for every LHS, i.e the $B$ matrix in $Ax=B$.If the homogeneous system has non-trivial solutions, then the inhomogeneous one is not solvable for every choice of $B$, and if it is solvable, it has infinitely many solutions.

But how can the inhomogeneous have infinitely many solutions, if it is solvable, I could't understand.

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    $\begingroup$ If $Ax = B$ and $Ay = 0$ for some $y \neq 0$, then $A(x + cy) = B$, for any scalar $c$ . $\endgroup$ Commented Jul 2, 2017 at 9:56
  • $\begingroup$ @DavidSchneider-Joseph you are right. I have never thought to consider the solution of the homogeneous one.Thanks a lot. $\endgroup$
    – Our
    Commented Jul 2, 2017 at 10:00
  • $\begingroup$ @DavidSchneider-Joseph By the way, if you post your comment as an answer, I will accept it. $\endgroup$
    – Our
    Commented Jul 2, 2017 at 10:01
  • $\begingroup$ As a nice follow-up exercise, see if you can prove that $\{x + y : Ay = 0\}$ contains all the solutions. $\endgroup$ Commented Jul 2, 2017 at 10:01

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If $Ax = B$ and $Ay = 0$ for some $y \neq 0$, then $A(x + cy) = B$, for any scalar $c$.

Conversely, suppose $Ax = B = Az$. Then define $y = z - x$, and we have that $z = x + y$ for some $y$ such that $Ay = 0$.

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