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Section 5.4 Expressing solutions to the equation \(A\xvec = \bvec\) in terms of the vector \(\xvec\)

Example 5.4.1.

We will describe the solution space of the equation
\begin{equation*} \left[\begin{array}{rrr} 2 \amp 0 \amp 2 \\ 4 \amp -1 \amp 6 \\ 1 \amp 3 \amp -5 \\ \end{array}\right] \xvec = \left[\begin{array}{r} 0 \\ -5 \\ 15 \end{array}\right]. \end{equation*}
This equation may be equivalently expressed as
\begin{equation*} x_1\left[\begin{array}{r}2\\4\\1\end{array}\right] + x_2\left[\begin{array}{r}0\\-1\\3\end{array}\right]+ x_3\left[\begin{array}{r}2\\6\\-5\end{array}\right]= \left[\begin{array}{r}0\\-5\\15\end{array}\right]\text{,} \end{equation*}
which is the linear system corresponding to the augmented matrix
\begin{equation*} \left[\begin{array}{rrr|r} 2 \amp 0 \amp 2 \amp 0 \\ 4 \amp -1 \amp 6 \amp -5 \\ 1 \amp 3 \amp -5 \amp 15 \\ \end{array} \right]\text{.} \end{equation*}
The reduced row echelon form of the augmented matrix is
\begin{equation*} \left[\begin{array}{rrr|r} 2 \amp 0 \amp 2 \amp 0 \\ 4 \amp -1 \amp 6 \amp -5 \\ 1 \amp 3 \amp -5 \amp 15 \\ \end{array} \right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp -2 \amp 5 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right], \end{equation*}
which corresponds to the linear system
\begin{equation*} \begin{alignedat}{4} x_1 \amp \amp \amp {}+{} \amp x_3 \amp {}={} \amp 0 \\ \amp \amp x_2 \amp {}-{} \amp 2x_3 \amp {}={} \amp 5. \\ \end{alignedat} \end{equation*}
The variable \(x_3\) is free so we may write the solution space parametrically as
\begin{equation*} \begin{aligned} x_1 \amp {}={} -x_3 \\ x_2 \amp {}={} 5+2x_3. \\ \end{aligned} \end{equation*}
Since we originally asked to describe the solutions to the equation \(A\xvec = \bvec\text{,}\) we will express the solution in terms of the vector \(\xvec\text{:}\)
\begin{equation*} \xvec =\left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{r} -x_3 \\ 5 + 2x_3 \\ x_3 \end{array} \right] =\left[\begin{array}{r}0\\5\\0\end{array}\right] +x_3\left[\begin{array}{r}-1\\2\\1\end{array}\right] \end{equation*}
As before, we call this a parametric description of the solution space.
This shows that the solutions \(\xvec\) may be written in the form \(\vvec + x_3\wvec\text{,}\) for appropriate vectors \(\vvec\) and \(\wvec\text{.}\) Geometrically, the solution space is a line in \(\real^3\) through \(\vvec\) moving parallel to \(\wvec\text{.}\)

Activity 5.4.1.

Consider the system of linear equations
\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp {}-{} \amp z \amp {}={} \amp 1 \\ 3x \amp {}+{} \amp 2y \amp {}+{} \amp 2z \amp {}={} \amp 7 \\ -x \amp \amp \amp {}+{} \amp 4z \amp {}={} \amp -3 \\ \end{alignedat}\text{.} \end{equation*}
  1. Find the matrix \(A\) and vector \(\bvec\) that expresses this linear system in the form \(A\xvec=\bvec\text{.}\)
  2. Give a description of the solution space to the equation \(A\xvec = \bvec\text{.}\)

Activity 5.4.2.

Consider the system of linear equations
\begin{equation*} \begin{alignedat}{4} x_1 \amp {}+{} \amp 2x_2 \amp {}+{} \amp x_3 \amp {}={} \amp 3 \\ 3x_1 \amp {}+{} \amp 2x_2 \amp {}+{} \amp x_3 \amp {}={} \amp 7 \\ -x_1 \amp \amp \amp \amp \amp {}={} \amp -2 \\ \end{alignedat}\text{.} \end{equation*}
  1. Find the matrix \(A\) and vector \(\bvec\) that expresses this linear system in the form \(A\xvec=\bvec\text{.}\)
  2. Give a description of the solution space to the equation \(A\xvec = \bvec\) in terms of the vector \(\xvec\text{.}\)
If \(A=\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\vvec_3 \end{array}\right]\) and \(\xvec=\left[ \begin{array}{c} x_1 \\ x_2 \\ x_3\\ \end{array}\right] \text{,}\) the following statements are equivalent.
  • The vector \(\xvec\) satisfies the equation \(A\xvec = \bvec \text{.}\)
  • The vector \(\bvec\) is a linear combination of the columns of \(A\) with weights \(x_j\text{:}\)
    \begin{equation*} x_1\vvec_1 + x_2\vvec_2 + x_3\vvec_3 = \bvec\text{.} \end{equation*}