# Vector space/Introductory example/Mulled wine/Example

At a booth on the Christmas market, there are three different pots of mulled wine. All three contain the ingredients cinnamon, cloves, red wine and sugar, but the compositions differ. The mixtures of the mulled wines are

${\displaystyle G_{1}={\begin{pmatrix}1\\2\\11\\2\end{pmatrix}},\,G_{2}={\begin{pmatrix}2\\2\\12\\3\end{pmatrix}},\,G_{3}={\begin{pmatrix}3\\1\\20\\7\end{pmatrix}}.}$

Every mulled wine is represented by a four-tuple, where the entries represent the respective shares of the ingredients. The set of all (possible) mulled wines form a vector space, and the three concrete mulled wines are vectors in this space.

Now suppose that none of the three mulled wines meets exactly our taste, in fact the wanted mulled wine has the mixture

${\displaystyle {}W={\begin{pmatrix}1\\2\\20\\5\end{pmatrix}}\,.}$

Is there a possibility to get the wanted mulled wine, by pouring together the given mulled wines in some way? Are there numbers[1] ${\displaystyle {}a,b,c\in \mathbb {Q} }$ such that

${\displaystyle {}a{\begin{pmatrix}1\\2\\11\\2\end{pmatrix}}+b{\begin{pmatrix}2\\2\\12\\3\end{pmatrix}}+c{\begin{pmatrix}3\\1\\20\\7\end{pmatrix}}={\begin{pmatrix}1\\2\\20\\5\end{pmatrix}}\,}$

holds? This vector-equation can be expressed by four equations in the "variables“ ${\displaystyle {}a,b,c}$, where the equations come from the rows. When does there exist a solution, when none, when many? These are typical questions of linear algebra.

1. In this example, only positive numbers have a practical interpretation. In linear algebra, everything is over a field, so we allow also negative numbers.