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The problem was: given the matrix $A \in M_{m\times n}$ and let $b \in F^{m}_{col}$. $y \in F^{m}_{col}$ is a solution of the system of equations $AX=b$. prove: every solution of the system of equations $AX=b$ can be represented as $y+x$, where $x \in F^{m}_{col}$ is the solution of the homogeneous system $AX=0$.

so my proof was like that: because $y \in F^{m}_{col}$ is a solution of the system of equations $AX=b$ we can conclude: $$(1) A \cdot y=b$$ and because $x \in F^{m}_{col}$ is a solution of the system of equations $AX=0$ we can conclude: $$(2) A \cdot x=0$$ from (1) and (2) we get: $$ A \cdot y+A \cdot x=0+b$$ according to the rules of matrix multiplication we get: $$(3) A \cdot ( y+ x)=0+b=b$$ and therefore according to (3) the solution which is represented by $z=x+y$ is also a solution of the system of equations $AX=b$

my instructor gave me only 3 points for that telling that "it isn't a valid proof". Why is that? what's the problem with it? and what should I say in order to appeal his decision. thank in advance.

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    $\begingroup$ You are only verifying that $y+x$ is a solution, not every solution is of the form $y+x$. $\endgroup$ Commented Feb 27, 2021 at 15:08

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This is the part that you missed: let $\;z\;$ be a solution, any solution, to $\;Ax=b\;$ , then

$$A(z-y)=Az-Ay=b-b=0\implies \;z-y\;\text{is a solution to the hom. system}\;\;Ax=0$$

and thus there exists $\;t\in\;$ space of all solutions to the hom. sytem, such that

$$z-y=t\implies z=t+y$$

with $\;t\;$ a solution to the hom. system, and $\;y\;$ a particular solution to the non-hom. system

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