Chapter 8 Review
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Chapter 8 Review
8-1
SYSTEMS OF LINEAR EQUATIONS IN TWO
VARIABLES; AUGMENTED MATRICES
a1 x1 b1 x2 k1
A system of two linear equations with two variables is a system
of the form
ax by h
cx dy k
(1)
where x and y are variables, a, b, c, and d are real numbers
called the coefficients of x and y, and h and k are real numbers
called the constant terms in the equations. The ordered pair of
numbers (x0, y0) is a solution to system (1) if each equation is
satisfied by the pair. The set of all such ordered pairs of numbers
is called the solution set for the system. To solve a system is to
find its solution set.
In general, a system of linear equations has exactly one solution, no solution, or infinitely many solutions. A system of linear equations is consistent if it has one or more solutions and
inconsistent if no solutions exist. A consistent system is said to
be independent if it has exactly one solution and dependent if
it has more than one solution.
Two standard methods for solving system (1) were reviewed: solution by graphing and solution using elimination
by addition.
Two systems of equations are equivalent if both have the
same solution set. A system of linear equations is transformed
into an equivalent system if:
a2 x1 b2 x2 k2
(2)
Associated with each linear system of the form (2), where x1 and
x2 are variables, is the augmented matrix of the system:
a
a1
b1
2
b2
Column 1 (C1)
Column 2 (C2)
Column 3 (C3)
k1
Row 1 (R1)
k2
Row 2 (R2)
(3)
Two augmented matrices are row-equivalent, denoted by
the symbol between the two matrices, if they are augmented
matrices of equivalent systems of equations. An augmented matrix is transformed into a row-equivalent matrix if any of the following row operations is performed:
1. Two rows are interchanged.
2. A row is multiplied by a nonzero constant.
3. A constant multiple of another row is added to a given row.
The following symbols are used to describe these row
operations:
1. Two equations are interchanged.
1. Ri ↔ Rj means “interchange row i with row j.”
2. An equation is multiplied by a nonzero constant.
2. kRi → Ri means “multiply row i by the constant k.”
3. A constant multiple of another equation is added to a given
equation.
3. kRj Ri → Ri means “multiply row j by the constant k and
add to Ri.”
These operations form the basis of solution using elimination by
addition. The method of solution using elimination by addition
can be transformed into a more efficient method for larger-scale
systems by the introduction of an augmented matrix. A matrix
is a rectangular array of numbers written within brackets. Each
number in a matrix is called an element of the matrix. If a matrix has m rows and n columns, it is called an m n matrix
(read “m by n matrix”). The expression m n is called the size
of the matrix, and the numbers m and n are called the dimensions of the matrix. A matrix with n rows and n columns is
called a square matrix of order n. A matrix with only one column is called a column matrix, and a matrix with only one row
is called a row matrix. The position of an element in a matrix
is the row and column containing the element. This is usually
denoted using double subscript notation aij, where i is the row
and j is the column containing the element aij.
For ease of generalization to larger systems, we change the
notation for variables and constants in system (1) to a subscript
form:
In solving system (2) using row operations, the objective is
to transform the augmented matrix (3) into the form
10
0 m
1 n
If this can be done, then (m, n) is the unique solution of system
(2). If (3) is transformed into the form
0
1
m n
0 0
then system (2) has infinitely many solutions. If (3) is transformed into the form
10
m n
0 p
p0
then system (2) does not have a solution.
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8-2
8 Systems of Equations and Inequalities
GAUSS=JORDAN ELIMINATION
In the last part of Section 8-1 we were actually using
Gauss–Jordan elimination to solve a system of two equations
with two variables. The method generalizes completely for systems with more than two variables, and the number of variables
does not have to be the same as the number of equations.
As before, our objective is to start with the augmented matrix of a linear system and transform it using row operations into
a simple form where the solution can be read by inspection. The
simple form, called the reduced form, is achieved if:
1. Each row consisting entirely of 0’s is below any row having
at least one nonzero element.
we investigated nonlinear systems involving second-degree
terms such as
x 2 y2 5
x 2 2y2 2
x 2 3x y y2 20
3x y 1
xy 2
x y y2 0
It can be shown that such systems have at most four solutions,
some of which may be imaginary.
Several methods were used to solve nonlinear systems of
the indicated form: solution by substitution, solution using
elimination by addition, and solution using factoring and
substitution. It is always important to check the solutions of
any nonlinear system to ensure that extraneous roots have not
been introduced.
2. The leftmost nonzero element in each row is 1.
3. The column containing the leftmost 1 of a given row has 0’s
above and below the 1.
4. The leftmost 1 in any row is to the right of the leftmost 1 in
the preceding row.
A reduced system is a system of linear equations that corresponds to a reduced augmented matrix. When a reduced system
has more variables than equations and contains no contradictions, the system is dependent and has infinitely many solutions.
The Gauss=Jordan elimination procedure for solving a
system of linear equations is given in step-by-step form as
follows:
Step 1. Choose the leftmost nonzero column, and use
appropriate row operations to get a 1 at the top.
Step 2. Use multiples of the row containing the 1 from
step 1 to get zeros in all remaining places in the
column containing this 1.
8-4
SYSTEMS OF LINEAR INEQUALITIES IN TWO
VARIABLES
A graph is often the most convenient way to represent the solution of a linear inequality in two variables or of a system of linear inequalities in two variables.
A vertical line divides a plane into left and right halfplanes. A nonvertical line divides a plane into upper and lower
half-planes. Let A, B, and C be real numbers with A and B not
both zero, then the graph of the linear inequality
Ax By C
or
Ax By C
with B 0, is either the upper half-plane or the lower half-plane
(but not both) determined by the line Ax By C. If B 0,
then the graph of
Ax C
or
Ax C
Step 3. Repeat step 1 with the submatrix formed by
(mentally) deleting the row used in step 2 and
all rows above this row.
is either the left half-plane or the right half-plane (but not both)
determined by the line Ax C. Out of these results follows an
easy step-by-step procedure for graphing a linear inequality
in two variables:
Step 4. Repeat step 2 with the entire matrix, including
the mentally deleted rows. Continue this
process until it is impossible to go further.
Step 1. Graph Ax By C as a broken line if equality is not included in the original statement or
as a solid line if equality is included.
If at any point in the above process we obtain a row with all 0’s
to the left of the vertical line and a nonzero number n to the
right, we can stop, since we have a contradiction: 0 n, n 0.
We can then conclude that the system has no solution. If this
does not happen and we obtain an augmented matrix in reduced
form without any contradictions, the solution can be read by
inspection.
Step 2. Choose a test point anywhere in the plane not
on the line and substitute the coordinates into
the inequality. The origin (0, 0) often requires
the least computation.
8-3
SYSTEMS INVOLVING SECOND-DEGREE
EQUATIONS
If a system of equations contains any equations that are not linear, then the system is called a nonlinear system. In this section
Step 3. The graph of the original inequality includes the
half-plane containing the test point if the
inequality is satisfied by that point, or the halfplane not containing that point if the inequality
is not satisfied by that point.
We now turn to systems of linear inequalities in two variables. The solution to a system of linear inequalities in two
variables is the set of all ordered pairs of real numbers that si-
Chapter 8 Review Exercise
multaneously satisfy all the inequalities in the system. The
graph is called the solution region. In many applications the solution region is also referred to as the feasible region. To find
the solution region, we graph each inequality in the system and
then take the intersection of all the graphs. A corner point of a
solution region is a point in the solution region that is the intersection of two boundary lines. A solution region is bounded if
it can be enclosed within a circle. If it cannot be enclosed within
a circle, then it is unbounded.
8-5
linear programming problem, and let z ax by be the objective function. If S is bounded, then z has both a maximum and a
minimum value on S and each of these occurs at a corner point
of S. If S is unbounded, then a maximum or minimum value of
z on S may not exist. However, if either does exist, then it must
occur at a corner point of S.
Problems with unbounded feasible regions are not considered in this brief introduction. The theorem leads to a simple
step-by-step solution to linear programming problems with
a bounded feasible region:
LINEAR PROGRAMMING
Linear programming is a mathematical process that has been
developed to help management in decision making, and it has
become one of the most widely used and best-known tools of
management science and industrial engineering.
A linear programming problem is one that is concerned
with finding the optimal value (maximum or minimum value)
of a linear objective function of the form z ax by, where
the decision variables x and y are subject to problem constraints in the form of linear inequalities and nonnegative constraints x, y 0. The set of points satisfying both the problem
constraints and the nonnegative constraints is called the feasible
region for the problem. Any point in the feasible region that
produces the optimal value of the objective function over the
feasible region is called an optimal solution. The fundamental
theorem of linear programming is basic to the solving of linear programming problems: Let S be the feasible region for a
655
Step 1. Form a mathematical model for the problem:
(A) Introduce decision variables and write a
linear objective function.
(B) Write problem constraints in the form of
linear inequalities.
(C) Write nonnegative constraints.
Step 2. Graph the feasible region and find the corner
points.
Step 3. Evaluate the objective function at each corner
point to determine the optimal solution.
If two corner points are both optimal solutions of the same
type (both produce the same maximum value or both produce
the same minimum value) to a linear programming problem,
then any point on the line segment joining the two corner points
is also an optimal solution of that type.
Chapter 8 Review Exercise
Work through all the problems in this chapter review and
check answers in the back of the book. Answers to all review
problems are there, and following each answer is a number
in italics indicating the section in which that type of problem
is discussed. Where weaknesses show up, review appropriate
sections in the text.
9. 2x y 2
x 2y 2
Perform each of the row operations indicated in Problems
10–12 on the following augmented matrix:
13
A
Solve Problems 1–6 using elimination by addition.
1. 2x y 7
3x 2y 0
3.
4x 3y 8
2x 32 y 4
5. x 2 y2 2
2x y 3
2.
3x 6y 5
2x 4y 1
4. y x 2 5x 3
y x 2
6. 3x 2 y2 6
2x 2 3y2 29
Solve Problems 7–9 by graphing.
7. 3x 2y 8
x 3y 1
8. 3x 4y 24
10. R1 ↔ R2
4
6
5
12
11. 13R2 → R2
12. (3)R1 R2 → R2
In Problems 13–15, write the linear system corresponding to
each reduced augmented matrix and solve.
13.
10
14.
10
1
0
15.
10
1
0
0
4
1 7
4
1
4
0