##### Example2.1.6

The previous activity also shows that questions about linear combinations lead naturally to linear systems. Let's ask how we can describe the vector $$\bvec=\left[\begin{array}{r} -1 \\ 4 \end{array} \right]$$ as a linear combination of $$\vvec$$ and $$\wvec\text{.}$$ We need to find weights $$a$$ and $$b$$ such that

\begin{equation*} \begin{aligned} a\left[\begin{array}{r}2\\1\end{array}\right] + b\left[\begin{array}{r}1\\2\end{array}\right] \amp = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \\ \left[\begin{array}{r}2a\\a\end{array}\right] + \left[\begin{array}{r}b\\2b\end{array}\right] \amp = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \\ \left[\begin{array}{r}2a+b\\a+2b\end{array}\right] \amp = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \end{aligned} \end{equation*}

Equating the components of the vectors on each side of the equation, we arrive at the linear system

\begin{equation*} \begin{alignedat}{3} 2a \amp {}+{} \amp b \amp {}={} \amp -1 \\ b \amp {}+{} \amp 2b \amp {}={} \amp 4 \\ \end{alignedat} \end{equation*}

This means that $$\bvec$$ is a linear combination of $$\vvec$$ and $$\wvec$$ if this linear system is consistent.

To solve this linear system, we construct its corresponding augmented matrix and find its reduced row echelon form.

\begin{equation*} \left[ \begin{array}{rr|r} 2 \amp 1 \amp -1 \\ 1 \amp 2 \amp 4 \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 3 \end{array} \right] \text{,} \end{equation*}

which tells us the weights $$a=-2$$ and $$b=3\text{;}$$ that is,

\begin{equation*} -2\vvec + 3 \wvec = \bvec \text{.} \end{equation*}

In fact, we know even more because the reduced row echelon matrix tells us that these are the only possible weights. Therefore, $$\bvec$$ may be expressed as a linear combination of $$\vvec$$ and $$\wvec$$ in exactly one way.

in-context