# If a corresponding homogeneous system of an inhomogeneous system has non-trivial solution, the inhomogeneous one has infinitely many solutions.How?

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.

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

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 $Ay = 0$.