# Question 4 A from Summer 2012 Exam Course 20109

Question 4

A) Let there be a non-homogenous set of linear equations in 4 variables such that the vectors, $u_1=(1,-2,4,3), u_2=(-3,1,2,1), u_3=(2,-3,3,-2), u_4=(1,3,-10,-4)$ are it's solutions. Find the general solution of this system.

Allot of us weren't sure how to answer this question, or even exactly what they were asking for. Here's a general outline of what I did, but I'm pretty sure it's mostly wrong.

The given implies that there is a matrix $A_{m\times4}$ and some vector $b\in\mathbb{F}^m$ such that $Au_i=b$ for each $u_i$.

I then checked that the given vectors where L.I. and then wrote a general solution as being any linear combination of the given vectors. I already know that's wrong because the zero vector isn't a solution.

So is anything I wrote correct, and what is the correct answer?

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You are correct that for some $A,b$ you have $A u_k = b$. Since it is non-homogeneous, we have $b\neq 0$, hence $\dim {\cal R}(A) \geq 1$. Since the dimension of the domain is $4$, this implies $\dim \ker A \leq 3$.
However, you are given that $A (u_k-u_1) = 0$, $k=2,3,4$, and a quick check shows that $u_k-u_1$, $k=2,3,4$ are linearly independent. Hence $\dim \ker A = 3$, and hence $\ker A = \text{sp} \{u_k-u_1\}_{k=2}^4$. Since $u_1$ is a particular solution, it follows that all solutions are of the form $u_1 + \sum_{k=2}^4 \alpha_k (u_k -u_1)$, where $\alpha_k \in \mathbb{R}$.