Example2.4.4

We will make the connection between solutions to the homogeneous equation and the linear independence of the columns more explict by looking at an example. In particular, we will demonstrate how a nontrivial solution to the homogeneous equation shows that one column of \(A\) is a linear combination of the others. With the matrix \(A\) in the previous activity, the homogeneous equation has the reduced row echelon form

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

which implies that

\begin{equation*} \begin{alignedat}{4} x_1 \amp \amp \amp {}+{} \amp 2x_3 \amp {}={} \amp 0 \\ \amp \amp x_2 \amp {}-{} \amp 3x_3 \amp {}={} \amp 0 \\ \end{alignedat} \text{.} \end{equation*}

In terms of the free variable \(x_3\text{,}\) we have

\begin{equation*} \begin{aligned} x_1 \amp {}={} -2x_3 \\ x_2 \amp {}={} 3x_3 \\ \end{aligned} \text{.} \end{equation*}

Any choice for a value of the free variable \(x_3\) produces a solution so let's choose, for convenience, \(x_3=1\text{.}\) We then have \((x_1,x_2,x_3) = (-2,3,1)\text{.}\)

Since \((-2,3,1)\) is a solution to the homogeneous equation \(A\xvec=\zerovec\text{,}\) this solution gives weights for a linear combination of the columns of \(A\) that create \(\zerovec\text{.}\) That is,

\begin{equation*} -2\vvec_1 + 3\vvec_2 + \vvec_3 = \zerovec \text{,} \end{equation*}

which we rewrite as

\begin{equation*} \vvec_3 = 2\vvec_1 - 3\vvec_2 \text{.} \end{equation*}
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