Sums of solutions of linear equations

Question

There's a question on an assignment that I'm not sure I quite understand:

Given a non-homogeneous system of equations $M$, determine for each of the following claims whether it is true or false:

1. If $\underline{c},\underline{d}$ are solutions of $M$, $(\underline{c}+\underline{d})$ is also a solution of $M$.

2. If $\underline{c},\underline{d}$ are solutions of $M$, $(3\underline{c}+2\underline{d})$ is also a solution of $M$.

Would the sum of solutions or multiples of those solutions be analogous to the sums of vectors containing the same values? I'm a little lost - some explanation would be appreciated.

EDIT:

I believe both statements are false... I've created a small, simple system to check the claims with:

$$ \begin{matrix}x+2y=1\\2x+4y=2\end{matrix}\Bigg\} \begin{bmatrix} 1&2&1\\2&4&2 \end{bmatrix}\vec{R_2 \rightarrow R_2-2 \cdot R_1}\begin{bmatrix}1&2&1\\0&0&0\end{bmatrix}\Rightarrow\Bigg\{\begin{matrix}x + 2y = 1\\0x+0y=0\end{matrix}\\\ \\ x+2y=1\Rightarrow x=1-2y\ ,\ y=y $$

Conclusion: the solution set for this system is: $$ S=\big\{(1-2y\ , \ y)\ |\ y\in R\big\} $$

Here are two possible specific solutions to the system:

$$ (-11\ ,\ 6)\ , \ (-17\ , \ 9)\in S $$

So if our first claim is correct that would mean:

$$ \big((-11\ ,\ 6)\ + \ (-17\ , \ 9)\big) \in S $$

Let's check:

$$ \big((-11\ , \ 6) + (-17\ , \ 9)\big) = (-28\ , \ 15)\\\ \\ (-28\ , \ 15)\in S \Rightarrow -28=1-2(15) $$

But since $-28\neq 1-2(15)$ the sum of the two solutions is not itself a solution of the system, so the first statement is false. Surely there are systems for which it is true, for example a system with the solution set

$$ T=\big\{(x,-x)\ |\ x\in R\big\} $$

the first statement is actually true.

I believe that by extension of this logic the second statement is also false.

Can someone confirm or deny my claims?

Answer 1

Both claim is not true in general.

say $c$ and $d$ are solutions to my non-homogeneous system of equation $Mx=y$ (so y is not a zero vector.)

so we have $Mc=y$ and $Md=y$

now $M(c+d) = 2y \neq y$ since $y$ is not a zero vector. So $(c+d)$ is not a solution.

I think you can see why $3c+2d$ is not a solution either.

Now if you are looking at a homogeneous system of equations, then statement 1 and 2 are ture.