I have to very simple question, that I am having some trouble with.
I am given an inhomogeneous linear system over $\mathbb{R}$ with four rows and three columns. We know that $v_p$ is a particular soloution.The question is then, is $2 \cdot v_p$ also a soloution?$v_p \quad expression$
I have already made some thoughts my self, but I would like to hear your thoughts.
EDIT:
Initially my thought was that $2 \cdot v_p$ is not a solution, because, if we have $A \cdot v_p = B$ we cant also have $A \cdot 2 \cdot v_p = B$
EDIT 2: It is a inhomogeneous, not homogeneous
$\quad$ There are only three cases for a linear systems of $Ax=b$: there are no solutions, there are unique solutions, and there are infinitely many solutions.
$\quad$ Since we already know a particular solution $v$, it can't be the first case. For the others, we can combine the discussion (if there is only a unique solution, then the subspace $X$ in the following article is interpreted as a $0$-dimensional subspace $\{0\}$).
$\quad$ So, you should know that the solution of $Ax=b$ has the range $X+a$, where $X$ is a subspace of $\mathbb R^n$($n$ is the number of columns in matrix $A$), and $X+a$ represents the set of every element in $X$ to which a constant $a$ is added. Here $a$ can be a particular solution for any of the linear systems. Since the linear systems is inhomogeneous, $a$ is not an element of $X$ (otherwise $X+a=X$, which means $0$ is a solution, which means that there must be $b=0$ in the linear systems $Ax=b$, which means the linear systems is homogeneous).
$\quad$ We know that for any element $x_0$ of a subspace $X$, $2x_0$ is also a member of that space. So if $v$ is an element of $X+a$, then there is an element $x$ of $X$ such that $x+a=v$, so $2v=2x+2a$. Suppose, $2v$ is also an element of $X+a$, then $2x+a$ must be an element of $X$, and since $2x$ is already an element of $X$, $a$ must be an element of $X$. There's a contradiction!
$\quad$ In summary, for a inhomogeneous linear systems with any row and any column, $2v$ must not be a solution as long as $v$ is a particular solution.