Solution set of a linear system with three equations and three unknowns with at least two distinct solutions.

Question

Proposition:

If $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ are two distinct solutions of a linear system with three equations and three unknowns then $t(x_1-x_0, y_1-y_0, z_1-z_0)$ is also a solution of that same system for any real number $t$.

I am uncertain on how I might go about determining the truth of this statement. It seems conceiveable, however, I can really only visualise this scenario for a line, rather than a plane. Things higher than $\mathbb{R}^2$ are rather difficult (or impossible) to conceive. It seems like subtracting these points gives a slop of some sort, and then $t$ could be considered the variable for the slope of some linear solution.

Insight?

Answer 1

Try to prove in not geometric but in algebraic way.

If you are familiar with matrices then express equations in matrix form and try to express $A \space t(x_1-x_0)$ in terms of $A \space x_1$ and $A \space x_0$. Then you will be able to easily prove the statement.

Answer 2

I think your question is not complete. It should mention homogenous equation somewhere. Lets think of the problem in Matrix notation. $$Ax = b$$ Where $A$ is a $3$ x $3$ Matrix and $b$ the solution vector. If $x^{(1)}=(x_1,y_1,z_1)$ and $x^{(0)}=(x_0,y_0,z_0)$ are solutions to your system, then lets define $u=t(x^{(1)}-x^{(0)})$. Plug this into the equation to get $$Au=A[t(x^{(1)}-x^{(0)})]=tAx^{(1)}-tAx^{(0)}$$ As $x^{(1)}$ and $x^{(0)}$ are solutions to the matrix equation $Ax^{(1)}=b$ and $Ax^{(0)}=b$. From $$Au=A[t(x^{(1)}-x^{(0)})]=tAx^{(1)}-tAx^{(0)}=t(b-b)=0$$ we conclude that $u$ is a solution to the homogenous equation.