Example 2.4.4

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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