Mathematics 209: Finite Mathematics

Study Guide

Unit 4: Systems of Linear Equations; Matrices

In this unit, you will learn two methods for solving systems of linear equations using matrices: Gauss-Jordan elimination and matrix equations.

Objectives

When you have completed this unit, you should be able to

1. solve a system of linear equations of two variables by substitution and by elimination by addition.
2. represent, graphically, the solution of a system of linear equations with two variables.
3. do elementary row operations in matrices.
4. solve a system of linear equations using the Gauss-Jordan elimination process.
5. do basic matrix operations.
6. obtain the inverse of an invertible matrix.
7. construct mathematical models that require a system of linear equations.

Review: Systems of Linear Equations in Two Variables

Indications

1. Read Section 4-1, Review: Systems of Linear Equations in Two Variables, on pages 178–187 of the textbook.
2. Read the Comments section below.
3. Do odd-numbered exercises 1–29 and 43–49 on pages 187–188 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
4. Solve odd-numbered problems 51–65 on pages 188–190 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
5. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Comments

A linear equation with two variables is an equation of the form

$ax+by=c,$

where $a,b,\; andc$ are any real numbers, and $xandy$ are variables.

The solution of a linear equation is a set of infinitely many numbers obtained by solving for one of the variables. If we solve for $y\text{,}$ then

${\displaystyle y=-\frac{ax}{b}+\frac{c}{b}.}$

We set $x=k$ for a parameter $k\text{,}$ and the solution set is all pairs of the form

${\displaystyle \left(k,-\frac{ak}{b}+\frac{c}{b}\right).}$

Example 4.1.  The solution set of the equation

$5x-7y=1$

consists of the pairs

${\displaystyle \left(k,\frac{5k}{7}-\frac{1}{7}\right).}$

Hence, for $k=7\text{,}$ we have the pair $\left(7,34∕7\right)$; and for $k=0\text{,}$ we have the pair $\left(0,1∕7\right)\text{.}$ Through these two points passes one and only one line, and this line is the solution of the linear equation. You should convince yourself that this line is equal to the given line.

A linear system of two variables corresponds to two lines in a two-dimensional space. If the lines are not parallel, then they intersect at one point. This point is the solution of the system. If the lines are parallel and not equal, then they do not intersect, and the system has no solution—it is “inconsistent.” If the lines are equal, then one is a multiple of the other, and either one is the solution of the system. Such a system has infinitely many solutions.

Two systems of linear equations are equivalent if they have the same solution.

Example 4.2.  In the system

$\begin{array}{cc}4x-6y\hfill & =2\hfill \\ 2x-3y\hfill & =1,\hfill \end{array}$

we see that the linear equations are multiples of each other, because $2\left(2x-3y\right)=1$ is equal to $4x-6y=2\text{.}$ Hence, the solution of this system is any one of these lines, say $2x-3y=1\text{.}$ As in the previous example, the solution consists of all pairs of the form

${\displaystyle \left(k,\frac{2k}{3}+\frac{1}{3}\right)}$

for the parameter $k\text{.}$ Thus, the system has infinitely many solutions.

Example 4.3.  In the system

$\begin{array}{cc}2x-3y\hfill & =1\hfill \\ 2x-3y\hfill & =-12\hfill \end{array}$

the slope of the lines is $2∕3\text{.}$ Hence, these lines are parallel. The $y$-intercept of the first line is $-1∕3\text{,}$ and that of the second line is $4\text{.}$ Hence, the lines are not equal and they do not intersect. This system has no solution—it is inconsistent.

Example 4.4.  In the system

$\begin{array}{cc}2x-3y\hfill & =1\hfill \\ 2x-5y\hfill & =-12\hfill \end{array}$

the slopes of the lines are $2∕3\text{,}$ and $2∕5\text{,}$ respectively. Hence, these lines intersect, and the solution is their point of intersection.

Questions

1. How exact is the graphing method used to solve a system of linear equations with two variables?
2. How do we know if a system of linear equations with two variables is inconsistent?
3. How do we know if a system of linear equations with two variables has only one solution?
4. When are two systems of linear equations equivalent?
5. How do we know if the mathematical model to solve a problem is a system of linear equations with two variables?

Systems of Linear Equations and Augmented Matrices

Indications

1. Read Section 4-2, Systems of Linear Equations and Augmented Matrices, on pages 190–198 of the textbook.
2. Read the Comments section below.
3. Do odd-numbered exercises 1–59 on pages 198–199 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
4. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Comments

In a system of three linear equations and two variables, we can represent three lines geometrically on a two-dimensional space, one line for each equation. The solution of the system is the set of points on the three lines. If the lines intersect at one point, the system has only one solution. If the lines do not intersect at any point, the system does not have a solution—it is inconsistent. If the lines intersect on one line, the system has infinitely many solutions.

The order in which the linear equations appear in the system is irrelevant to the solution. The lines remain the same, which means that the solution is the same. So, if we interchange the linear equations in a system, the resulting system is equivalent to the original one.

We indicated in the previous section that a line remains the same if we multiply it by a constant. Therefore, when we multiply the linear equations in a system by a constant, the resulting system is equivalent to the original one.

If we have a system of linear equations

$\begin{array}{cc}{a}_{1}x+b_{1}y\hfill & =c_1\hfill \\ a_{2}x+b_{2}y\hfill & =c_2\hfill \\ a_{3}x+b_{3}y\hfill & =c_3\hfill \end{array}$

(4.1)

and we multiply any of the linear equations by a constant $k$ and add it to another equation, then

$\begin{array}{cc}{a}_{1}x+b_{1}y\hfill & =c_1\hfill \\ a_{2}x+b_{2}y\hfill & =c_2\hfill \\ \left(k{a}_1+a_3\right)x+\left(k{b}_1+b_3\right)y\hfill & =k{c}_1+c_3\hfill \end{array}$

(4.2)

If $\left(a,b\right)$ is a solution of System (4.1), then

$\begin{array}{cc}{a}_{1}a+b_{1}b\hfill & =c_1\hfill \\ a_{2}a+b_{2}b\hfill & =c_2\hfill \\ a_{3}a+b_{3}b\hfill & =c_3,\hfill \end{array}$

and

$\begin{array}{cc}k{a}_{1}a+k{b}_{1}b\hfill & =k{c}_1\hfill \\ a_{2}a+b_{2}b\hfill & =c_2\hfill \\ a_{3}a+b_{3}b\hfill & =c_3.\hfill \end{array}$

Thus,

$\begin{array}{cc}k{a}_{1}a+k{b}_{1}b\hfill & =k{c}_1\hfill \\ a_{2}a+b_{2}b\hfill & =c_2\hfill \\ a_{3}a+b_{3}b+k{c}_1\hfill & =k{c}_1+c_3,\hfill \end{array}$

and we have

$\begin{array}{cc}k{a}_{1}a+k{b}_{1}b\hfill & =k{c}_1\hfill \\ a_{2}a+b_{2}b\hfill & =c_2\hfill \\ k{a}_{1}a+a_{3}a+k{b}_{1}b+b_{3}b\hfill & =k{c}_1+c_3\hfill \end{array}$

Therefore, this system is equivalent to the one below

$\begin{array}{cc}{a}_{1}a+b_{1}b\hfill & =c_1\hfill \\ a_{2}a+b_{2}b\hfill & =c_2\hfill \\ \left(k{a}_1+a_3\right)a+\left(k{b}_1+b_3\right)b\hfill & =k{c}_1+c_3.\hfill \end{array}$

This result shows that $\left(a,b\right)$ is also a solution of system (4.2). Since both systems have the same solutions, they are equivalent.

We have shown with these observations that the row operations listed in Theorem 1 on page 193 of the textbook yield an equivalent system of equations.

Questions

6. What are the dimensions of the augmented matrix that corresponds to a system of four linear equations of two variables?
7. How do we know if a system of three linear equations and two variables has only one solution?
8. Why do we perform row operations on an augmented matrix to solve a system of linear equations?
9. If an augmented matrix is of the form

   $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right],$

   what is the solution of the system?

10. If an augmented matrix is of the form

    $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right],$

    with $a\ne 0\text{,}$ what is the solution of the system?

Gauss-Jordan Elimination

Indications

1. Read Section 4-3, Gauss-Jordan Elimination, on pages 199–208 of the textbook.
2. Read the Comments section below.
3. Follow along with the three tutorials in the Tutorials section below.
4. Do odd-numbered exercises 1–59 on pages 209–211 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
5. Solve odd-numbered problems 63–89 on pages 211–213 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
6. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Comments

In a matrix, the first nonzero number in a row is called the pivot.

Example 4.5.  In the matrix

$\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 0.8\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill -5\hfill & \hfill -1\hfill \\ \hfill 6\hfill & \hfill 0\hfill & \hfill 0.1\hfill \end{array}\right],$

the pivot of the first row is $4\text{,}$ that of the second row is $-5$ and that of the third row is $6\text{.}$

A matrix in “reduced row echelon form” is a matrix such that

1. the rows consisting of zeros are below any row with one nonzero entry.
2. each pivot is a $1\text{.}$
3. there are zeros above and below each pivot.
4. if there is a pivot in column $A$ of a row, then the pivot of the row below is in a column to the right of column $A\text{.}$

Example 4.6.  The matrices

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0.8\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]B=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

are not in reduced form. In the second row of matrix $A\text{,}$ the pivot has a nonzero number, $0.8\text{,}$ above it. In matrix $B\text{,}$ the pivot in the first row is in the second column, while the pivot in the second row is in the first column, which is to the left, not the right, of the second column.

Example 4.7.  The matrices

$A=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]B=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

are in reduced form. [Why?]

There is more than one way to obtain the reduced form of a matrix. Take the procedure on page 203 of the textbook as a recommendation. In the following example, we present a different strategy to obtain the reduced form of Example 3 on pages 203–204.

Example 4.8.  The augmented matrix of the system is

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 4\hfill & \hfill -8\hfill & \hfill 7\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 4\hfill & \hfill -3\hfill & \hfill 5\hfill \end{array}\right].$

We are trying to avoid ending up with fractions, so we will not multiply the first row by $1∕2\text{,}$ but we will try to obtain zeros in the first column.

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 4\hfill & \hfill -8\hfill & \hfill 7\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 4\hfill & \hfill -3\hfill & \hfill 5\hfill \end{array}\right]\left(-2\right)R_1+R_2\to R_2$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 10\hfill \\ \hfill -2\hfill & \hfill 4\hfill & \hfill -3\hfill & \hfill 5\hfill \end{array}\right]R_1+R_3\to R_3$

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 10\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 1\hfill \end{array}\right]\left(\frac{1}{5}\right)R_2}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 1\hfill \end{array}\right]\left(2\right)R_2+R_3\to R_3$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill \end{array}\right]$

The solution given in the textbook stops here, but we will continue, to show you how we obtain the reduced form.

Now that Condition 4 (p. 34 of this Study Guide) is satisfied, we can make the pivot of the third row a $1\text{,}$ and use it to make zeros above it.

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill \end{array}\right]\left(\frac{1}{5}\right)R_3}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left(-2\right)R_3+R_2\to R_2$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left(4\right)R_3+R_1\to R_1$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Now we make zeros above the pivot of the second row:

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left(-1\right)R_2+R_1\to R_1$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Finally, we make the first pivot of the first row a $1$:

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill -4\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left(\frac{1}{2}\right)R_1}$

$\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Example 4.9.  See Exercise 50 on page 210.

The augmented matrix is

$\left[\begin{array}{cccc}\hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill & \hfill -10\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \end{array}\right].$

We will avoid fractions again—you must follow the operations carefully to see how.

$\left[\begin{array}{cccc}\hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill & \hfill -10\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \end{array}\right]R_3\to R_1$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill & \hfill -10\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \end{array}\right]\left(3\right)R_1;\left(-2\right)R_2$

$\left[\begin{array}{cccc}\hfill 6\hfill & \hfill 12\hfill & \hfill -3\hfill & \hfill -3\hfill \\ \hfill -6\hfill & \hfill 2\hfill & \hfill 4\hfill & \hfill 20\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \end{array}\right]R_1+R_2\to R_2$

${\displaystyle \left[\begin{array}{cccc}\hfill 6\hfill & \hfill 12\hfill & \hfill -3\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \end{array}\right]\left(\frac{1}{3}\right)R_1}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 3\hfill \end{array}\right]\left(-2\right)R_1+R_3\to R_3$

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 0\hfill & \hfill -10\hfill & \hfill 5\hfill & \hfill 5\hfill \end{array}\right]\left(\frac{7}{5}\right)R_3}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 0\hfill & \hfill -14\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right]R_2+R_3\to R_3$

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 8\hfill & \hfill 24\hfill \end{array}\right]\left(\frac{1}{8}\right)R_3}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 1\hfill & \hfill 17\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]-R_3+R_2\to R_2$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 0\hfill & \hfill 14\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]R_3+R_1\to R_1$

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 0\hfill & \hfill 14\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\left(\frac{1}{14}\right)R_2}$

$\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\left(-4\right)R_2+R_1\to R_1$

${\displaystyle \left[\begin{array}{cccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\left(\frac{1}{2}\right)R_1}$

${\displaystyle \left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]}$

The solution is $x_1=-1\text{,}$ $x_2=1$, and $x_3=3\text{.}$

If we replace these values in the system of equations, we obtain

$4\left(-1\right)-2\left(1\right)+3\left(3\right)=3$

$3\left(-1\right)-\left(1\right)-2\left(3\right)=-10$

$2\left(-1\right)+4\left(1\right)-\left(3\right)=-1$

This check shows that the answer is correct.

Note: The reduced form of a matrix is unique; that is, it does not matter which row operations we use to find the reduced form, we will always end up with the same reduced form.

Consider Example 4 on page 202 of the textbook. Observe that the reduced form does not have a pivot in the third column, and that the general solution has a parameter in the third variable; that is, $x_3=t\text{.}$ Similarly, in Example 5 on page 204, the reduced form does not have a pivot in the third and fifth columns, and the variables $x_3$ and $x_5$ correspond to parameters; that is, $x_3=s$ and $x_5=t\text{.}$

Tutorials

These presentation slides with audio explanations should help you gain confidence with some of the more complex topics in this section.

• The Pivot Element
• Strategies to Obtain the Reduced Echelon Form
• Solving a System of Linear Equations
• This tutorial explains the echelon form, which can be used to solve a system of linear equations when obtaining the reduced echelon form poses a challenge.

Questions

11. How do we know from the reduced form of an augmented matrix that a system is inconsistent?
12. How do we know from the reduced form of an augmented matrix that a system has only one solution?
13. How do we know from the reduced form of an augmented matrix that a system has infinitely many solutions?

Matrices: Basic Operations

To solve a matrix equation we need to learn to do basic operations (sum and product) with matrices.

Indications

1. Read Section 4-4, Matrices: Basic Operations, on pages 213–222 of the textbook.
2. Read the Comments section below.
3. Follow along with the tutorial in the Tutorials section below.
4. Do odd-numbered exercises 1–57 on pages 222–223 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
5. Solve odd-numbered problems 61–69 on pages 224–226 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
6. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Comments

If a matrix is of dimensions $m\times n\text{,}$ then the entry (number) in the $i$-th row and $j$-th column is denoted by $a_{ij}\text{.}$

If the matrices $A$ and $B$ have dimensions $m\times n$ and $n\times p\text{,}$ respectively, then their product $AB$ has dimensions $m\times p\text{.}$ The $a_{ij}$ entry of the product matrix $AB$ is the product of the $i$-th row of the matrix $A$ and the $j$-th column of the matrix $B\text{.}$ For instance, the entry $a_{23}$ is the product of the second row of $A$ times the third column of $B\text{.}$

Example 4.10.  Consider the following matrices

$A=\left[\begin{array}{cccc}\hfill 4\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill -2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right]B=\left[\begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill -2\hfill & \hfill -1\hfill & \hfill 6\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill -3\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill -1\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right].$

The product $AB$ has dimensions $3\times 5\text{.}$ The entry $a_{24}$ is the product of the second row of $A$ and the fourth column of $B\text{,}$ thus

$a_{24}=\left[\begin{array}{cccc}\hfill 3\hfill & \hfill 1\hfill & \hfill -2\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -1\hfill \\ \hfill 1\hfill \\ \hfill 3\hfill \end{array}\right]=0-1-2-3=-6$

The $a_{35}$ entry is

$a_{35}=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 5\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]=3+30+0-1=32$

Tutorials

These presentation slides with audio explanations should help you gain confidence with some of the more complex topics in this section.

• Multiplication of Matrices

Questions

14. If $A$ and $B$ are matrices of the same dimensions, is it true that $A+B=B+A$ and that $A+A+A=3A\text{?}$ Explain.
15. If $A$ and $B$ are square matrices, is it true that $AB=BA\text{?}$ Explain.
16. If $A$ and $B$ are square matrices, is it true that if $AB=0\text{,}$ then $A=0$ or $B$ = 0? Explain.

    Note: Here $0$ is the zero matrix; that is, all entries are $0\text{.}$

17. If $A$ and $B$ are square matrices, is it true that if $AB=0\text{,}$ then $BA=0\text{?}$ Explain.

    Note: Here $0$ is the zero matrix; that is, all entries are $0\text{.}$

Inverse of a Square Matrix

To solve the equation $3x=12\text{,}$ we multiply it by the inverse of 3, that is ${\displaystyle \frac{1}{3}}$; hence

${\displaystyle \frac{1}{3}\left(3x\right)=\frac{1}{3}12\text{ and }1x=4,}$

and we conclude that $x=4\text{.}$ Two properties allow us to reach this conclusion:

1. multiplication by $1$; that is, for any number $x\text{,}$ we have

   $1x=x.$

2. use of the inverse; for example, for the number $3\text{,}$ there is a number ${\displaystyle \frac{1}{3}\text{,}}$ the inverse of $3\text{,}$ such that

   ${\displaystyle \frac{1}{3}\left(3\right)=1.}$

Similarly, to solve a matrix equation $AX=C\text{,}$ where $A\text{,}$ $X\text{,}$ and $C$ are matrices, we need an identity $I$; that is, a matrix such that $AI=IA=A\text{.}$ We also need an inverse of $A$; that is, a matrix $A^{-1}$ such that $A^{-1}A=I\text{.}$ If we have both, then

$A^{-1}\left(AX\right)=A^{-1}CIX=A^{-1}CX=A^{-1}C.$

In this section, you will learn which matrices have the two properties identified above.

Indications

1. Read Section 4-5, Inverse of a Square Matrix, on pages 227–236 of the textbook.
2. Do odd-numbered exercises 1–59 on pages 236–237 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
3. Solve problems 65 and 67 on page 237 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
4. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Questions

18. Which matrices are invertible?

    Hint: See Theorem 1 on page 230.

19. If $R$ and $S$ are two rows in a square matrix $M$ and $R=kS$ for some constant $k\text{,}$ is the matrix $M$ invertible?
20. Is it true that if a matrix $M$ is invertible, then it has only one inverse $M^{-1}\text{?}$ Explain.
21. Is it true that if a matrix $M$ is invertible, then its inverse matrix $M^{-1}$ is also invertible?
22. Why do we consider only square matrices when we look for the inverse of a matrix?

Matrix Equations and Systems of Linear Equations

Indications

1. Read Section 4-6, Matrix Equations and Systems of Linear Equations, on pages 238–244 of the textbook.
2. Do odd-numbered exercises 1–43 on pages 244–245 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
3. Solve problems 49 and 51 on page 246 of the textbook. If you have difficulty, consult your tutor to discuss the problem.
4. Answer the questions listed below, and then compare your answers with those given in the Answers to Study Guide Questions.

Questions

23. Why do we have a left and a right distributive property for matrices?
24. Is it true that for any nonzero matrices $A\text{,}$ $B\text{,}$ $C\text{,}$ if $AB=AC\text{,}$ then $B=C\text{?}$ Explain.
25. Can you think of any nonzero matrices $A\text{,}$ $B\text{,}$ $C\text{,}$ such that if $AB=AC\text{,}$ then $B=C\text{?}$ Explain.

Finishing This Unit

1. Review the objectives of this unit and make sure you are able to meet each of them.
2. Study the section of the Chapter 4 Review titled Important Terms, Symbols and Concepts, on pages 255–256 of the textbook.
3. If there is a concept, definition, example or exercise that is not yet clear to you, go back and re-read it. Contact your tutor if you need help.
4. You might want to do odd-numbered exercises 1–37 and 41–47 from the Review Exercise section on pages 256–259 of the textbook. The questions on the practice examination are taken from this exercise. If you have difficulty, consult your tutor to discuss the problem.
5. Complete the practice examination provided for this unit. Evaluate yourself, first checking your answers against those provided in the Answers section at the end of the textbook, and then comparing your solutions with those provided on pages S-205 to S-225 of the Student Solutions Manual portion of the textbook. The number of points in a question may indicate the number of steps in the solution. Give yourself full credit if your answer is correct and you give a complete solution, even if your solution differs from that shown in the Student Solutions Manual.
6. Once you have written and evaluated the practice examination, complete and submit the second assignment. You will find instructions in the Assignment Drop Box on the course home page. Also, please leave enough time to enable your tutor to return your marked assignment with feedback before you write your examination.

Practice Examination

Time: 1.5 hours
Total points: 44
Passing grade: 50%

Do the following exercises from the Chapter 4 Review Exercise on pages 257–259 of the textbook.

To obtain full credit you must justify all your answers and show your work.

1. Exercise 24  (Marks: 6 pts )
2. Exercise 26, part (C)  (Marks: 6 pts )
3. Exercise 34  (Marks: 8 pts.)
4. Exercise 38  (Marks: 6 pts.)
5. Exercise 44  (Marks: 12 pts.)
6. Exercise 46  (Marks: 6 pts.)