TITLE: Don't understand what this question about solutions of linear equations is asking.
QUESTION [5 upvotes]: If there exists a solution for the linear equation $A x = \left( \begin{array}{lll} 1 \\ 1 \\ 1 \end{array} \right)$, then there also exists a solution for $Ax = \left( \begin{array}{lll} 1 \\ 2 \\ 1 \end{array} \right)$.

I have a list of statements, and I need to prove or disprove them. 
Here is an example of such a statement. 
I don't understand the format of the question. 
I know a contradictory example to the statement I'm attaching is a square matrix $3 \times 3$ of all $1$'s. Why?

REPLY [4 votes]: Here’s what this particular question is asking:

Suppose that $A$ is a $3\times 3$ matrix such that the equation $$Ax=\left[\matrix{1\\1\\1}\right]\tag{1}$$ has a solution; is it then necessarily true that the equation $$Ax=\left[\matrix{1\\2\\1}\right]\tag{2}$$ also has a solution?

The answer is no, because when $$A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\;,$$ equation $(1)$ has a solution, but equation $(2)$ does not.
$$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{1\\0\\0}\right]=\left[\matrix{1\\1\\1}\right]\;,$$ but 
$$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{x\\y\\z}\right]=\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\;,$$ and no matter what $x,y$, and $z$ you choose, $$\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\ne\left[\matrix{1\\2\\1}\right]\;,$$ because all three of its components are the same.

In general, if your asked to prove or disprove a statement of the form $$\text{if }P\text{ is true then }Q\text{ is true}\;,$$ you first have to decide whether it is true: if the mathematical objects mentioned in the statement are chosen in such a way that $P$ is true, does that automatically make $Q$ true as well? If the answer is yes, you’ll need to find a reason, a logical argument leading from the truth of $P$ to the truth of $Q$. For example, suppose that the statement is:

if the $3\times 3$ matrix $A$ is invertible, then the equation $$Ax=\left[\matrix{a\\b\\c}\right]$$ has a solution for every possible choice of real numbers $a,b,c$,

you can show that it’s true by pointing to the solution $$x=Ix=(A^{-1}A)x=A^{-1}(Ax)=A^{-1}\left[\matrix{a\\b\\c}\right]\;.$$
If the statement is not always true, your job is in some ways easier: you just have to demonstrate one example in which $P$ is true and $Q$ is not. That’s what I did in the specific example that you gave: the matrix $$A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]$$ shows that equation $(1)$ can have a solution without $(2)$ also having a solution.