Given two vectors $$\vec{u}$$ and $$\vec{v}$$ we name linear combination of $$\vec{u}$$ and $$\vec{v}$$ to any expression of the form: $$\lambda\vec{u}+\mu\vec{v}$$ where $$\lambda$$ and $$\mu$$ are real numbers.

A vector $$\vec{w}$$ is a linear combination of $$\vec{u}$$ and $$\vec{v}$$ if real (scalar) numbers (escalars) $$\lambda$$ and $$\mu$$ exist such that we can express $$\vec{w}$$ as follows: $$\vec{w}=\lambda\vec{u}+\mu\vec{v}$$.

The vectors we have been working with until now are vectors on the plane, so they have two components. In this case we can express any vector $$\vec{w}$$ as a linear combination of two non parallel vectors $$\vec{u}$$ and $$\vec{v}$$. This combination is unique.

Is the vector $$\vec{w}=(-1,3)$$ a linear combination of the vectors of $$\vec{u}=(1,2)$$ and $$\vec{v}=(0,3)$$?

We want to find $$\lambda$$ and $$\mu$$ so as $$\vec{w}= \lambda\vec{u}+\mu\vec{v}$$. We have: $$$ (-1,3)=\lambda(1,2)+\mu(0,3)= (\lambda,2\lambda)+(0,3\mu)= (\lambda, 2\lambda+3\mu)$$$

Therefore: $$$\left. \begin{array}{rcl} -1&=&\lambda \\ 3&=&2\lambda+3\mu \end{array} \right\} \Rightarrow \lambda=-1, \ \mu=\dfrac{5}{3}$$$

We have just found values for $$\lambda$$ and $$\mu$$ for which $$\vec{w}= \lambda\vec{u}+\mu\vec{v}$$ is true. So the answer is "yes", we can express $$\vec{w}=(-1,3)$$ as a linear combination of $$\vec{u}=(1,2)$$ and $$\vec{v}=(0,3)$$.