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Section 1.3 Solving Linear Systems (E3)

Activity 1.3.1.

Free browser-based technologies for mathematical computation are available online.

  • Go to https://sagecell.sagemath.org/.

  • Type A=Matrix([[1,3,4],[2,5,7]]) to store the matrix \(\left[\begin{array}{} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]\) in the variable \(A\text{.}\) Then add the line show(A) and use the Evaluate button to preview it.

  • Add the line show(A.rref()) and use Evaluate to compute the reduced row echelon form of \(A\text{.}\)

Since the vertical bar in an augmented matrix does not affect row operations, the \(\RREF\) of \(\left[\begin{array}{cc|c} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]\) may be computed in the same way.

Activity 1.3.2.

Consider the following system of equations.

\begin{align*} 3x_1 &\,-\,& 2x_2 &\,+\,& 13x_3 &\,=\,& 6\\ 2x_1 &\,-\,& 2x_2 &\,+\,& 10x_3 &\,=\,& 2\\ -x_1 &\,+\,& 3x_2 &\,-\,& 6x_3 &\,=\,& 11 \end{align*}
(a)

Convert this to an augmented matrix and use technology to compute its reduced row echelon form:

\begin{equation*} \RREF \left[\begin{array}{ccc|c} \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \end{array}\right] = \left[\begin{array}{ccc|c} \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \end{array}\right] \end{equation*}
(b)

Use the \(\RREF\) matrix to write a linear system equivalent to the original system. Then find its solution set.

Activity 1.3.3.

Consider the vector equation

\begin{equation*} x_1 \left[\begin{array}{c} 3 \\ 2\\ -1 \end{array}\right] +x_2 \left[\begin{array}{c}-2 \\ -2 \\ 0 \end{array}\right] +x_3\left[\begin{array}{c} 13 \\ 10 \\ -3 \end{array}\right] =\left[\begin{array}{c} 6 \\ 2 \\ 1 \end{array}\right] \end{equation*}
(a)

Convert this to an augmented matrix and use technology to compute its reduced row echelon form:

\begin{equation*} \RREF \left[\begin{array}{ccc|c} \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \end{array}\right] = \left[\begin{array}{ccc|c} \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \unknown&\unknown&\unknown&\unknown\\ \end{array}\right] \end{equation*}
(b)

Use the \(\RREF\) matrix to write a linear system equivalent to the original system. Then find its solution set.

Activity 1.3.4.

Consider the following linear system.

\begin{alignat*}{4} x_1 &+ 2x_2 &+ 3x_3 &= 1\\ 2x_1 &+ 4x_2 &+ 8x_3 &= 0 \end{alignat*}
(a)

Find its corresponding augmented matrix \(A\) and use technology to find \(\RREF(A)\text{.}\)

(b)

How many solutions do these linear systems have?

Activity 1.3.5.

Consider the simple linear system equivalent to the system from the previous activity:

\begin{alignat*}{3} x_1 &+ 2x_2 & &= 4\\ & &\phantom{+}x_3 &= -1 \end{alignat*}
(a)

Let \(x_1=a\) and write the solution set in the form \(\setBuilder { \left[\begin{array}{c} a \\ \unknown \\ \unknown \end{array}\right] }{ a \in \IR } \text{.}\)

(b)

Let \(x_2=b\) and write the solution set in the form \(\setBuilder { \left[\begin{array}{c} \unknown \\ b \\ \unknown \end{array}\right] }{ b \in \IR } \text{.}\)

(c)

Which of these was easier? What features of the RREF matrix \(\left[\begin{array}{ccc|c} \circledNumber{1} & 2 & 0 & 4 \\ 0 & 0 & \circledNumber{1} & -1 \end{array}\right]\) caused this?

Definition 1.3.1.

Recall that the pivots of a matrix in \(\RREF\) form are the leading \(1\)s in each non-zero row.

The pivot columns in an augmented matrix correspond to the bound variables in the system of equations (\(x_1,x_3\) below). The remaining variables are called free variables (\(x_2\) below).

\begin{equation*} \left[\begin{array}{ccc|c} \circledNumber{1} & 2 & 0 & 4 \\ 0 & 0 & \circledNumber{1} & -1 \end{array}\right] \end{equation*}

To efficiently solve a system in RREF form, assign letters to the free variables, and then solve for the bound variables.

Activity 1.3.6.

Find the solution set for the system

\begin{alignat*}{6} 2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\ -x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\ x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2 \end{alignat*}

by row-reducing its augmented matrix, and then assigning letters to the free variables (given by non-pivot columns) and solving for the bound variables (given by pivot columns) in the corresponding linear system.

Observation 1.3.2.

The solution set to the system

\begin{alignat*}{6} 2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\ -x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\ x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2 \end{alignat*}

may be written as

\begin{equation*} \setBuilder { \left[\begin{array}{c} 1+5a+2b \\ 1+2a+3b \\ a \\ 3+3b \\ b \end{array}\right] }{ a,b\in \IR }\text{.} \end{equation*}
Remark 1.3.3.

Don't forget to correctly express the solution set of a linear system. Systems with zero or one solutions may be written by listing their elements, while systems with infinitely-many solutions may be written using set-builder notation.

  • Consistent with one solution: e.g. \(\setList{ \left[\begin{array}{c}1\\2\\3\end{array}\right] }\)

  • Consistent with infinitely-many solutions: e.g. \(\setBuilder { \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] }{ a\in\IR }\)

  • Inconsistent: \(\emptyset\) or \(\{\}\)

Exercises 1.3.1 Exercises

1.

Show how to find the solution set for the following system of linear equations.

\begin{alignat*}{5} x_{1} &-& 2 \, x_{2} &-& x_{3} &+& 3 \, x_{4} &=& 4\\ 2 \, x_{1} &-& 3 \, x_{2} &-& 3 \, x_{3} &+& 5 \, x_{4} &=& 6\\ &-& 3 \, x_{2} &+& 3 \, x_{3} &+& 3 \, x_{4} &=& 6 \end{alignat*}
Answer
\begin{equation*} \mathrm{RREF} \left[\begin{array}{cccc|c} 1 & -2 & -1 & 3 & 4 \\ 2 & -3 & -3 & 5 & 6 \\ 0 & -3 & 3 & 3 & 6 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & -3 & 1 & 0 \\ 0 & 1 & -1 & -1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

The solution set is \(\left\{ \left[\begin{array}{c} 3 \, a - b \\ a + b - 2 \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)

2.

Show how to find the solution set for the following system of linear equations.

\begin{alignat*}{4} x_{1} & & &-& x_{3} &=& 4\\ x_{1} &+& x_{2} & & &=& 3\\ x_{1} &+& x_{2} &+& x_{3} &=& -2\\ & & & & x_{3} &=& -4 \end{alignat*}
Answer
\begin{equation*} \mathrm{RREF} \left[\begin{array}{ccc|c} 1 & 0 & -1 & 4 \\ 1 & 1 & 0 & 3 \\ 1 & 1 & 1 & -2 \\ 0 & 0 & 1 & -4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \end{equation*}

The solution set is \(\left\{\right\} \)

3.

Show how to find the solution set for the following system of linear equations.

\begin{alignat*}{4} 4 \, x_{1} &+& x_{2} &+& 7 \, x_{3} &=& 6\\ -5 \, x_{1} &-& 4 \, x_{2} &-& 6 \, x_{3} &=& -2\\ -x_{1} &+& x_{2} &-& 3 \, x_{3} &=& -4\\ -3 \, x_{1} & & &-& 6 \, x_{3} &=& -6 \end{alignat*}
Answer
\begin{equation*} \mathrm{RREF} \left[\begin{array}{ccc|c} 4 & 1 & 7 & 6 \\ -5 & -4 & -6 & -2 \\ -1 & 1 & -3 & -4 \\ -3 & 0 & -6 & -6 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 2 & 2 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

The solution set is \(\left\{ \left[\begin{array}{c} -2 \, a + 2 \\ a - 2 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)

4.

Show how to find the solution set for the following system of linear equations.

\begin{alignat*}{4} x_{1} &-& x_{2} &-& x_{3} &=& -6\\ & & x_{2} &-& 3 \, x_{3} &=& 4\\ & & & & 0 &=& 1\\ & & 2 \, x_{2} &-& 6 \, x_{3} &=& 7 \end{alignat*}
Answer
\begin{equation*} \mathrm{RREF} \left[\begin{array}{ccc|c} 1 & -1 & -1 & -6 \\ 0 & 1 & -3 & 4 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & -6 & 7 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -4 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

The solution set is \(\left\{\right\} \)

5.

Show how to find the solution set for the following vector equation

\begin{equation*} x_{1} \left[\begin{array}{c} -3 \\ 0 \\ -1 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ -1 \end{array}\right] + x_{3} \left[\begin{array}{c} 7 \\ -5 \\ 3 \\ 7 \end{array}\right] = \left[\begin{array}{c} 1 \\ 6 \\ 0 \\ -2 \end{array}\right] . \end{equation*}
Answer
\begin{equation*} \mathrm{RREF} \left[\begin{array}{ccc|c} -3 & -1 & 7 & 1 \\ 0 & 1 & -5 & 6 \\ -1 & 0 & 3 & 0 \\ -2 & -1 & 7 & -2 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

The solution set is \(\left\{ \left[\begin{array}{c} -3 \\ 1 \\ -1 \end{array}\right] \right\} \)

Additional exercises available at checkit.clontz.org.