# General solution of a system of equations given a set of specific solutions

I'm pretty sure I did this right but I'd just like to check to make sure - I've been presented with the following problem:

Given a non-homogenous system of equations with 4 variables so that the vectors: $(0,1,2,3),(1,2,3,4),(1,2,4,5),(5,8,3,1)$ are solutions to the system, find the general solution of the system of equations.

My reasoning was as follows: the solution set to the system contains those four vectors, meaning if I set them up as rows of a matrix and use row operations to get it to reduced row echelon form, I should at least have an indicator as to the dimension of the solution set and go from there. The thing is, when I did just that I was left with an identity matrix, meaning the solution set spans $\mathbb{R}^4$. With that in mind, I answered that the solution set (which I called $B$) is $B=\{(x_1,x_2,x_3,x_4)\ |\ x_1...x_4\in \mathbb{R}\}$.

The reason I believe this is true is that if I have four solution vectors which are, at the end of the day, four linearly independent vectors in a four-dimensional vector space ($\mathbb{R}^4$), then any vector in that space could be made using linear combinations of those four vectors. It follows that the solution set spans the entire vector space and thus, any vector in that space could be a solution to the system.

On the other hand this seems weird to me and I feel like there might be a flaw in my logic somewhere, because there can't possibly be a system of equations to which any vector in $\mathbb{R}^4$ could be a solution, could there?

• Do you said linear equations? Commented Feb 3, 2016 at 22:27
• Yes, the system is linear. Commented Feb 3, 2016 at 22:34

You have a system $$A = \begin{pmatrix}a_{11}&a_{12}&a_{13}&a_{14} \\a_{11}&a_{12}&a_{13}&a_{14} \\a_{11}&a_{12}&a_{13}&a_{14}\\a_{11}&a_{12}&a_{13}&a_{14} \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix}$$
where at least one of the $b_i;\space i=1,2,3,4$ is non zero (since non-homogenous).
Your conclusion is correct, since your four solutions generate all $\mathbb R^4$.