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.