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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/.

  • In the dropdown on the right, you can select a number of different languages. Select "Octave" for the Matlab-compatible syntax used by this text.

  • Type rref([1,3,2;2,5,7]) and then press the Evaluate button to compute the \(\RREF\) of \(\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]\text{.}\)

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.

In the HTML version of this text, code cells are often embedded for your convenience when RREFs need to be computed.

Try this out to compute \(\RREF\left[\begin{array}{cc|c} 2 & 3 & 1 \\ 3 & 0 & 6 \end{array}\right]\text{.}\)

Activity 1.3.3.

Consider the following system of equations.

\begin{alignat*}{4} 3x_1 &\,-\,& 2x_2 &\,+\,& 13x_3 &\,=\,& 6\\ 2x_1 &\,-\,& 2x_2 &\,+\,& 10x_3 &\,=\,& 2\\ -x_1 &\,+\,& 3x_2 &\,-\,& 6x_3 &\,=\,& 11\text{.} \end{alignat*}

(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 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 1.3.5.

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 1.3.6.

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 1.3.7.

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 1.3.8.

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 1.3.9.

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 1.3.10.

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 \(\{\}\)

Subsection 1.3.1 Videos

Figure 1.3.11. Video: Solving a system of linear equations with infinitely-many solutions

Exercises 1.3.2 Exercises

Exercises available at checkit.clontz.org 1 .

https://checkit.clontz.org/#/banks/tbil-la/E3/