Download 8.5 Determinants and Cramer`s Rule

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856 Chapter 8 Matrices and Determinants
90. If A = B
3
2
Preview Exercises
5
-1
R , find 1A-12 .
4
91. Find values of a for which the following matrix is not invertible:
B
1
a - 2
a + 1
R.
4
Exercises 93–95 will help you prepare for the material covered in
the next section. Simplify the expression in each exercise.
93. 21-52 - 1-32142
94.
Group Exercise
92. Each person in the group should work with one partner.
Send a coded word or message to each other by giving your
partner the coded matrix and the coding matrix that you
selected. Once messages are sent, each person should decode
the message received.
Section
8.5
21- 52 - 11- 42
51- 52 - 61- 42
95. 21- 30 - 1- 322 - 316 - 92 + 1 - 1211 - 152
Determinants and Cramer’s Rule
Objectives
A portion of Charles Babbage’s unrealized
Difference Engine
� Evaluate a second-order
�
�
�
�
�
�
determinant.
Solve a system of linear
equations in two variables
using Cramer’s rule.
Evaluate a third-order
determinant.
Solve a system of linear
equations in three variables
using Cramer’s rule.
Use determinants to identify
inconsistent systems and
systems with dependent
equations.
Evaluate higher-order
determinants.
Evaluate a second-order
determinant.
s cyberspace absorbs more and
more of our work, play, shopping,
and socializing, where will it all end?
Which activities will still be offline in
2025?
Our technologically transformed
lives can be traced back to the English
inventor Charles Babbage (1792–1871).
Babbage knew of a method for
solving linear systems called Cramer’s rule, in honor of the Swiss geometer Gabriel
Cramer (1704–1752). Cramer’s rule was simple, but involved numerous
multiplications for large systems. Babbage designed a machine, called the
“difference engine,” that consisted of toothed wheels on shafts for performing
these multiplications. Despite the fact that only one-seventh of the functions ever
worked, Babbage’s invention demonstrated how complex calculations could be
handled mechanically. In 1944, scientists at IBM used the lessons of the difference
engine to create the world’s first computer.
Those who invented computers hoped to relegate the drudgery of repeated
computation to a machine. In this section, we look at a method for solving linear
systems that played a critical role in this process. The method uses real numbers,
called determinants, that are associated with arrays of numbers. As with matrix
methods, solutions are obtained by writing down the coefficients and constants of a
linear system and performing operations with them.
A
The Determinant of a 2 : 2 Matrix
Associated with every square matrix is a real number, called its determinant. The
determinant for a 2 * 2 square matrix is defined as follows:
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Section 8.5 Determinants and Cramer’s Rule
Study Tip
Definition of the Determinant of a 2 : 2 Matrix
To
evaluate
a
second-order
determinant, find the difference of the
product of the two diagonals.
`
a1
b1
a2
b2
857
` = a1b2 - a2b1
The determinant of the matrix B
`
a1
a2
b1
a
R is denoted by ` 1
b2
a2
b1
` and is defined by
b2
b1
` = a1b2 - a2b1 .
b2
a1
a2
We also say that the value of the second-order determinant `
a1
a2
b1
` is
b2
a1b2 - a2b1 .
Example 1 illustrates that the determinant of a matrix may be positive or
negative. A determinant can also have 0 as its value.
Evaluating the Determinant of a 2 : 2 Matrix
EXAMPLE 1
Evaluate the determinant of each of the following matrices:
a. B
Discovery
Write and then evaluate three
determinants, one whose value is
positive, one whose value is negative,
and one whose value is 0.
5
7
6
R
3
b. B
a. `
5 "6
" ` = 5 # 3 - 7 # 6 = 15 - 42 = - 27
7 3
b. `
2
-3
a. B
Solve a system of linear
equations in two variables
using Cramer’s rule.
4
R.
-5
Solution We multiply and subtract as indicated.
10
6
The value of the secondorder determinant is -27.
"4
" ` = 21- 52 - 1-32142 = - 10 + 12 = 2
-5
Check Point
�
2
-3
1
The value of the secondorder determinant is 2.
Evaluate the determinant of each of the following matrices:
9
R
5
b. B
4
-5
3
R.
-8
Solving Systems of Linear Equations
in Two Variables Using Determinants
Determinants can be used to solve a linear system in two variables. In general, such
a system appears as
b
a1x + b1y = c1
a2x + b2y = c2 .
Let’s first solve this system for x using the addition method. We can solve for x by
eliminating y from the equations. Multiply the first equation by b2 and the second
equation by -b1 . Then add the two equations:
b
a1x + b1y = c1
a2x + b2y = c2
Multiply by b2 .
"
Multiply by - b1 .
b
"
Add:
a1b2x + b1b2y = c1b2
-a2b1x - b1b2y = - c2b1
1a1b2 - a2b12x =
x =
c1b2 - c2b1
c1b2 - c2b1
.
a1b2 - a2b1
Because
`
c1
c2
b1
` = c1b2 - c2b1
b2
and
`
a1
a2
b1
` = a1b2 - a2b1 ,
b2
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858 Chapter 8 Matrices and Determinants
we can express our answer for x as the quotient of two determinants:
x =
`
c1
c2
c1b2 - c2b1
=
a1b2 - a2b1
a
` 1
a2
b1
`
b2
b1
`
b2
.
Similarly, we could use the addition method to solve our system for y, again expressing y
as the quotient of two determinants. This method of using determinants to solve the
linear system, called Cramer’s rule, is summarized in the box.
Solving a Linear System in Two Variables Using Determinants
Cramer’s Rule
If
b
a1x + b1y = c1
a2x + b2y = c2,
then
`
x =
c1
c2
b1
`
b2
a
` 1
a2
b1
`
b2
y =
and
`
a1
a2
c1
`
c2
a
` 1
a2
b1
`
b2
,
where
`
a1
a2
b1
` Z 0.
b2
Here are some helpful tips when solving
b
a1x + b1y = c1
a2x + b2y = c2
using determinants:
1. Three different determinants are used to find x and y. The determinants in the
denominators for x and y are identical. The determinants in the numerators
for x and y differ. In abbreviated notation, we write
x =
Dx
D
and y =
Dy
D
, where D Z 0.
2. The elements of D, the determinant in the denominator, are the coefficients of
the variables in the system.
a
b1
D = ` 1
`
a2 b2
3. Dx , the determinant in the numerator of x, is obtained by replacing the
x-coefficients, in D, a1 and a2 , with the constants on the right sides of the
equations, c1 and c2 .
D = `
a1
a2
b1
`
b2
and
Dx = `
c1
c2
b1
`
b2
Replace the column with a1 and a2 with
the constants c 1 and c2 to get Dx .
4. Dy , the determinant in the numerator for y, is obtained by replacing the
y-coefficients, in D, b1 and b2 , with the constants on the right sides of the
equations, c1 and c2 .
D = `
a1
a2
b1
`
b2
and
Dy = `
a1
a2
c1
`
c2
Replace the column with b1 and b2 with
the constants c1 and c2 to get Dy .
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Section 8.5 Determinants and Cramer’s Rule
859
Using Cramer’s Rule to Solve a Linear System
EXAMPLE 2
Use Cramer’s rule to solve the system:
b
5x - 4y = 2
6x - 5y = 1.
Solution Because
x =
Dx
D
and
Dy
y =
D
,
we will set up and evaluate the three determinants D, Dx , and Dy .
1. D, the determinant in both denominators, consists of the x- and y-coefficients.
D = `
5
6
-4
` = 1521-52 - 1621- 42 = -25 + 24 = -1
-5
Because this determinant is not zero, we continue to use Cramer’s rule to solve
the system.
2. Dx , the determinant in the numerator for x, is obtained by replacing the
x-coefficients in D, 5 and 6, by the constants on the right sides of the equations,
2 and 1.
Dx = `
2
1
-4
` = 1221-52 - 1121- 42 = -10 + 4 = -6
-5
3. Dy , the determinant in the numerator for y, is obtained by replacing the
y-coefficients in D, -4 and - 5, by the constants on the right sides of the
equations, 2 and 1.
Dy = `
5
6
2
` = 152112 - 162122 = 5 - 12 = -7
1
4. Thus,
x =
Dx
-6
=
= 6
D
-1
and y =
Dy
D
=
-7
= 7.
-1
As always, the solution (6, 7) can be checked by substituting these values into
the original equations. The solution set is 516, 726.
Check Point
2
Use Cramer’s rule to solve the system:
e
3
Evaluate a third-order
determinant.
5x + 4y = 12
3x - 6y = 24.
The Determinant of a 3 : 3 Matrix
Associated with every square matrix is a real number called its determinant. The
determinant for a 3 * 3 matrix is defined as follows:
Definition of a Third-Order Determinant
a1
3 a2
a3
b1
b2
b3
c1
c2 3 = a1b2c3 + b1c2a3 + c1a2b3 - a3b2c1 - b3c2a1 - c3a2b1
c3
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The six terms and the three factors in each term in this complicated evaluation formula,
a1b2c3 + b1c2a3 + c1a2b3 - a3b2c1 - b3c2a1 - c3a2b1, can be rearranged, and then
we can apply the distributive property. We obtain
a1b2c3 - a1b3c2 - a2b1c3 + a2b3c1 + a3b1c2 - a3b2c1
= a11b2c3 - b3c22 - a21b1c3 - b3c12 + a31b1c2 - b2c12
= a1 `
b2
b3
c2
b
` - a2 ` 1
c3
b3
c1
b
` + a3 ` 1
c3
b2
c1
`.
c2
You can evaluate each of the second-order determinants and obtain the three
expressions in parentheses in the second step.
In summary, we now have arranged the definition of a third-order determinant
as follows:
Definition of the Determinant of a 3 : 3 Matrix
A third-order determinant is defined by
Subtract.
a1
3 a2
a3
Add.
b1 c1
b c
b c
b c
b2 c2 3 =a1 2 2 2 2 – a2 2 1 1 2 +a3 2 1 1 2 .
b3 c3
b3 c3
b2 c2
b3 c3
Each a on the right comes
from the first column.
Here are some tips that may be helpful when evaluating the determinant of a
3 * 3 matrix:
Evaluating the Determinant of a 3 : 3 Matrix
1. Each of the three terms in the definition contains two factors—a numerical
factor and a second-order determinant.
2. The numerical factor in each term is an element from the first column of the
third-order determinant.
3. The minus sign precedes the second term.
4. The second-order determinant that appears in each term is obtained by
crossing out the row and the column containing the numerical factor.
a1 `
b2 c2
b c
b c
` - a2 ` 1 1 ` + a3 ` 1 1 `
b3 c3
b3 c3
b2 c2
a1 b1 c1 a1 b1 c1 a1 b1 c1
3 a2 b2 c2 3 3 a2 b2 c2 3 3 a2 b2 c2 3
a3 b3 c3 a3 b3 c3 a3 b3 c3
The minor of an element is the determinant that remains after deleting the
row and column of that element. For this reason, we call this method
expansion by minors.
EXAMPLE 3
Evaluating the Determinant of a 3 : 3 Matrix
Evaluate the determinant of the following matrix:
4
C -9
-3
1
3
8
0
4 S.
1
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Section 8.5 Determinants and Cramer’s Rule
861
Solution We know that each of the three terms in the determinant contains a
numerical factor and a second-order determinant. The numerical factors are
from the first column of the given matrix. They are highlighted in the following
matrix:
4
C -9
-3
1
3
8
0
4 S.
1
We find the minor for each numerical factor by deleting the row and column of that
element:
4
C –9
–3
1
3
8
0
4S
1
4
C –9
–3
The minor for
4 is 3 4 .
8 1
1
3
8
0
4S
1
4
C –9
–3
The minor for
−9 is 1 0 .
8 1
兩 兩
1
3
8
0
4S
1
The minor for
−3 is 1 0 .
3 4
兩 兩
兩 兩
Now we have three numerical factors, 4, - 9, and -3, and three second-order
determinants. We multiply each numerical factor by its second-order determinant
to find the three terms of the third-order determinant:
4`
Technology
A graphing utility can be used to
evaluate the determinant of a matrix.
Enter the matrix and call it A. Then
use the determinant command. The
screen below verifies our result in
Example 3.
3
8
4
`,
1
-9 `
0
`,
1
1
8
-3 `
1
3
0
`.
4
Based on the preceding definition, we subtract the second term from the first term
and add the third term:
Don't forget to
supply the minus sign.
4
3 –9
–3
=
=
=
=
=
1
3
8
0
4 3 =4 2 3 4 2 -(–9) 2 1 0 2 – 3 2 1 0 2 Begin by evaluating the three
8 1
8 1
3 4 second-order determinants.
1
413 # 1 - 8 # 42 + 911 # 1 - 8 # 02 - 311 # 4 - 3 # 02
Multiply within parentheses.
413 - 322 + 911 - 02 - 314 - 02
Subtract within parentheses.
41- 292 + 9112 - 3142
Multiply.
- 116 + 9 - 12
Add and subtract as indicated.
- 119
Check Point
3
Evaluate the determinant of the following matrix:
2
C -5
-4
1
6
3
7
0 S.
1
The six terms in the definition of a third-order determinant can be rearranged
and factored in a variety of ways.Thus, it is possible to expand a determinant by minors
about any row or any column. Minus signs must be supplied preceding any element
appearing in a position where the sum of its row and its column is an odd number. For
example, expanding about the elements in column 2 gives us
a1
3 a2
a3
b1 c1
a c
a c
a c
b2 c2 3=–b1 2 2 2 2+b2 2 1 1 2-b3 2 1 1 2 .
a3 c3
a3 c3
a2 c2
b3 c3
Minus sign is supplied because
b1 appears in row 1 and column 2;
1 + 2 = 3, an odd number.
Minus sign is supplied because
b3 appears in row 3 and column 2;
3 + 2 = 5, an odd number.
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862 Chapter 8 Matrices and Determinants
Expanding by minors about column 3, we obtain
Study Tip
a1
3 a2
a3
Keep in mind that you can expand a
determinant by minors about any row
or column. Use alternating plus and
minus signs to precede the numerical
factors of the minors according to the
following sign array:
+
3+
+
-
+
- 3.
+
b1 c1
a b
a b
a b
b2 c2 3=c1 2 2 2 2-c2 2 1 1 2+c3 2 1 1 2 .
a3 b3
a3 b3
a2 b2
b3 c3
Minus sign must be supplied because
c2 appears in row 2 and column 3;
2 + 3 = 5, an odd number.
When evaluating a 3 * 3 determinant using expansion by minors, you can
expand about any row or column. To simplify the arithmetic, if a row or column
contains one or more 0s, expand about that row or column.
Evaluating a Third-Order Determinant
EXAMPLE 4
Evaluate:
9
3 -2
1
5
-3
4
0
03.
2
Solution Note that the last column has two 0s. We will expand the determinant
about the elements in that column.
9
3 -2
1
5
-3
4
0
-2
03 = 0`
1
2
-3
9
` - 0`
4
1
5
9
` + 2`
4
-2
= 0 - 0 + 2391- 32 - 1- 2254
= 21- 27 + 102
= 21- 172
= - 34
Check Point
4
Solve a system of linear
equations in three variables
using Cramer’s rule.
Evaluate the second-order
determinant whose numerical
factor is not 0.
Evaluate:
6
3 -3
1
�
5
`
-3
4
-5
2
0
33.
0
Solving Systems of Linear Equations
in Three Variables Using Determinants
Cramer’s rule can be applied to solving systems of linear equations in three variables.
The determinants in the numerator and denominator of all variables are third-order
determinants.
Solving Three Equations in Three Variables Using Determinants
Cramer’s Rule
If
a1x + b1y + c1z = d1
c a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
then
x =
Dy
Dz
Dx
,y =
, and z =
, where D Z 0.
D
D
D
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Section 8.5 Determinants and Cramer’s Rule
863
These four third-order determinants are given by
a1
D = 3 a2
a3
b1
b2
b3
c1
c2 3 These are the coefficients of the variables x, y, and z.
c3
d1
3
Dx = d2
d3
b1
b2
b3
c1
Replace x-coefficients in D with the constants on the right
c2 3 of the three equations.
c3
a1
Dy = 3 a2
a3
d1
d2
d3
c1
Replace y-coefficients in D with the constants on the right
c2 3 of the three equations.
c3
a1
3
Dz = a2
a3
b1
b2
b3
d1
Replace z-coefficients in D with the constants on the right
d2 3 of the three equations.
d3
EXAMPLE 5
Using Cramer’s Rule to Solve
a Linear System in Three Variables
Use Cramer’s rule to solve:
x + 2y - z = -4
c x + 4y - 2z = -6
2x + 3y + z = 3.
Solution Because
Dy
Dz
Dx
, y =
, and z =
,
D
D
D
we need to set up and evaluate four determinants.
Step 1 Set up the determinants.
1. D, the determinant in all three denominators, consists of the x-, y-, and
z-coefficients.
1 2 -1
D = 3 1 4 -2 3
x =
2 3
1
2. Dx , the determinant in the numerator for x, is obtained by replacing the
x-coefficients in D, 1, 1, and 2, with the constants on the right sides of the
equations, -4, - 6, and 3.
-4 2 - 1
3
Dx = - 6 4 -2 3
3 3
1
3. Dy , the determinant in the numerator for y, is obtained by replacing the
y-coefficients in D, 2, 4, and 3, with the constants on the right sides of the
equations, -4, - 6, and 3.
1 -4 - 1
3
Dy = 1 -6 - 2 3
2
3
1
4. Dz , the determinant in the numerator for z, is obtained by replacing the
z-coefficients in D, -1, -2, and 1, with the constants on the right sides of the
equations, -4, - 6, and 3.
1 2 -4
3
Dz = 1 4 - 6 3
2 3
3
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864 Chapter 8 Matrices and Determinants
Step 2 Evaluate the four determinants.
1
D=3 1
2
2 –1
4
4 –2 3 =1 2
3
3 1
2
–2
2 -1 2
3
1
2
–1
2 ±22
4
1
–1
2
–2
= 114 + 62 - 112 + 32 + 21-4 + 42
= 11102 - 1152 + 2102 = 5
Study Tip
To find Dx , Dy , and Dz , you’ll need
to apply the evaluation process for a
3 * 3 determinant three times. The
values of Dx , Dy , and Dz cannot be
obtained from the numbers that
occur in the computation of D.
Using the same technique to evaluate each determinant, we obtain
Dx = - 10, Dy = 5, and
Dz = 20.
Step 3 Substitute these four values and solve the system.
x =
y =
z =
Dx
- 10
=
= -2
D
5
Dy
D
=
5
= 1
5
=
20
= 4
5
Dz
D
The solution 1-2, 1, 42 can be checked by substitution into the original three
equations. The solution set is 51 - 2, 1, 426.
Check Point
5
Use Cramer’s rule to solve the system:
3x - 2y + z = 16
c 2x + 3y - z = - 9
x + 4y + 3z = 2.
�
Use determinants to identify
inconsistent systems and
systems with dependent
equations.
Cramer’s Rule with Inconsistent and Dependent Systems
If D, the determinant in the denominator, is 0, the variables described by the
quotient of determinants are not real numbers. However, when D = 0, this
indicates that the system is either inconsistent or contains dependent equations. This
gives rise to the following two situations:
Determinants: Inconsistent and Dependent Systems
1. If D = 0 and at least one of the determinants in the numerator is not 0, then
the system is inconsistent. The solution set is ⭋.
2. If D = 0 and all the determinants in the numerators are 0, then the equations
in the system are dependent. The system has infinitely many solutions.
Discovery
Write a system of two equations that is inconsistent. Now use determinants and the result
boxed above to verify that this is truly an inconsistent system. Repeat the same process for a
system with two dependent equations.
Although we have focused on applying determinants to solve linear systems,
they have other applications, some of which we consider in the exercise set that
follows Example 6.
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Section 8.5 Determinants and Cramer’s Rule
�
Evaluate higher-order
determinants.
865
The Determinant of Any n : n Matrix
The determinant of a matrix with n rows and n columns is said to be an nth-order
determinant. The value of an nth-order determinant 1n 7 22 can be found in terms
of determinants of order n - 1. For example, we found the value of a third-order
determinant in terms of determinants of order 2.
We can generalize this idea for fourth-order determinants and higher. We
have seen that the minor of the element aij is the determinant obtained by deleting
the ith row and the jth column in the given array of numbers. The cofactor of the
element aij is 1-12i + j times the minor of aij. If the sum of the row and
column 1i + j2 is even, the cofactor is the same as the minor. If the sum of the row and
column 1i + j2 is odd, the cofactor is the opposite of the minor.
Let’s see what this means in the case of a fourth-order determinant.
Evaluating the Determinant of a 4 : 4 Matrix
EXAMPLE 6
Evaluate the determinant of the following matrix:
1
-1
A = D
0
2
-2
1
2
3
3
0
0
-4
0
2
T.
-3
1
Solution
1
-1
ƒAƒ = 4
0
2
-2
1
2
3
3
0
0
-4
–1
=(–1)1 ± 3(3) 3 0
2
0
24
-3
1
1 2
1 –2 0
2 –3 3+(–1)4 ± 3(–4) 3 –1 1 2 3
3 1
0 2 –3
−4 is in row 4,
column 3.
3 is in row 1,
column 3.
-1
= 33 0
2
1
2
3
With two 0s in the third column, we
will expand along the third column.
2
1
-3 3 + 4 3 -1
1
0
-2
1
2
The determinant that follows 3 is
obtained by crossing out the row and
the column (row 1, column 3) in the
original determinant. The minor for - 4
is obtained in a similar manner.
0
23
-3
Evaluate the two third-order determinants to get
ƒ A ƒ = 31- 252 + 41- 12 = - 79.
Check Point
6
Evaluate the determinant of the following matrix:
0
-1
A = D
1
3
4
1
-2
0
0
5
0
0
-3
2
T.
6
1
If a linear system has n equations, Cramer’s rule requires you to compute
n + 1 determinants of nth order. The excessive number of calculations required
to perform Cramer’s rule for systems with four or more equations makes it an
inefficient method for solving large systems.
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866 Chapter 8 Matrices and Determinants
Exercise Set 8.5
Practice Exercises
Evaluate each determinant in Exercises 1–10.
1. `
5
2
3. `
-4
5
5. `
7
`
3
8
`
6
2. `
4
5
1
`
6
4. `
7
-2
9
`
-5
-7
2
14
`
-4
6. `
1
-8
-3
`
2
7. `
-5
-2
-1
`
-7
8. `
1
5
1
6
-6
5
9. `
1
2
1
8
1
2
3
4
`
10. `
2
3
1
2
1
3
3
4
-
-
4x - 5y - 6z = - 1
35. c x - 2y - 5z = - 12
2x - y
=
7
x - 3y + z = - 2
36. c x + 2y
= 8
2x - y
= 1
x + y + z = 4
37. c x - 2y + z = 7
x + 3y + 2z = 4
2x + 2y + 3z = 10
38. c 4x - y + z = - 5
5x - 2y + 6z = 1
x +
`
3x + 2z = 4
40. c 5x - y = - 4
4y + 3z = 22
2z = 4
2y - z = 5
2x + 3y
= 13
39. c
Evaluate each determinant in Exercises 41–44.
`
For Exercises 11–26, use Cramer’s rule to solve each system or
to determine that the system is inconsistent or contains
dependent equations.
11. b
x + y = 7
x - y = 3
12. b
2x + y = 3
x - y = 3
13. b
12x + 3y = 15
2x - 3y = 13
14. b
x - 2y = 5
5x - y = - 2
4
-2
41. 4
5
4
2
0
0
0
-2
1
43. 4
1
2
-3
-4
2
0
8
4
0
0
-7
14
5
-1
3
0
2
1
5
04
-3
1
3
-2
42. 4
2
1
-1
0
-1
4
1
0
-2
2
2
04
3
3
1
3
44. 4
2
2
-3
-1
1
0
2
0
3
-2
0
-2 4
1
0
0
`
-3
-5
`
6
`
Practice Plus
In Exercises 45–46, evaluate each determinant.
4x - 5y = 17
15. b
2x + 3y = 3
3x + 2y = 2
16. b
2x + 2y = 3
`
1
`
3
0
`
7
`
`
x + 2y = 3
5x + 10y = 15
18. b
2x - 9y = 5
3x - 3y = 11
19. b
3x - 4y = 4
2x + 2y = 12
20. b
3x = 7y + 1
2x = 3y - 1
In Exercises 47–48, write the system of linear equations for which
Cramer’s rule yields the given determinants.
21. b
2x = 3y + 2
5x = 51 - 4y
22. b
y = - 4x + 2
2x = 3y + 8
47. D = `
2
3
-4
`,
5
Dx = `
8
-10
23. b
3x = 2 - 3y
2y = 3 - 2x
24. b
x + 2y - 3 = 0
12 = 8y + 4x
48. D = `
2
5
-3
`,
6
Dx = `
8
11
25. b
4y = 16 - 3x
6x = 32 - 8y
26. b
2x = 7 + 3y
4x - 6y = 3
In Exercises 49–52, solve each equation for x.
3
3
27. 2
2
0
1
5
0
-5 3
-1
4
3
28. 3
2
3
29. 3 - 3
-1
1
4
3
0
03
-5
2
30. 3 - 1
3
1
31. 3 2
-3
1
2
4
1
23
-5
1
32. 3 2
3
0
-1
-3
0
43
5
-4
0
0
2
2
2
-2
4
1
51. 3 3
0
2
53
4
3
-3 3
1
In Exercises 33–40, use Cramer’s rule to solve each system.
x + y + z = 0
33. c 2x - y + z = - 1
-x + 3y - z = - 8
49. `
x - y + 2z = 3
34. c 2x + 3y + z = 9
- x - y + 3z = 11
5
4
46. 4
7
`
4
x
` = 32
6
x
1
-2
-4
`
5
-3
`
6
50. `
x + 3
x - 2
2
52. 3 - 3
2
-2
1 3 = -8
2
-1
0
4
`
-3
0
`
-1 4
1
`
5
17. b
Evaluate each determinant in Exercises 27–32.
7
1
9
`
3
0
`
5 4
-6
`
5
3
-2
45. 4
3
`
0
x
1
1
-6
` = 28
-4
1
0 3 = 39
4
Application Exercises
Determinants are used to find the area of a triangle whose vertices
are given by three points in a rectangular coordinate system. The
area of a triangle with vertices 1x1 , y12, 1x2 , y22, and 1x3 , y32 is
Area = ;
x
13 1
x2
2
x3
y1
y2
y3
1
13,
1
where the ; symbol indicates that the appropriate sign should
be chosen to yield a positive area. Use this information to work
Exercises 53–54.
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Section 8.5 Determinants and Cramer’s Rule
53. Use determinants to find the area of the triangle whose
vertices are 13, - 52, (2, 6), and 1- 3, 52.
54. Use determinants to find the area of the triangle whose
vertices are (1, 1), 1 - 2, -32, and 111, -32.
Determinants are used to show that three points lie on the same
line (are collinear). If
x1
3 x2
x3
y1
y2
y3
1
1 3 = 0,
1
then the points 1x1 , y12, 1x2 , y22, and 1x3 , y32 are collinear. If the
determinant does not equal 0, then the points are not collinear. Use
this information to work Exercises 55–56.
867
In Exercises 68–69, use a graphing utility to evaluate the
determinant for the given matrix.
3
-5
68. D
2
-1
-2
1
4
3
-1
2
5
-6
8
2
69. E 2
-1
4
4
7
T
0
5
2
0
1
2
5
6
-3
-3
1
-2
-1
4
6
5
3
0
7
-5 U
-1
-8
70. What is the fastest method for solving a linear system with
your graphing utility?
Critical Thinking Exercises
55. Are the points 13, - 12, 10, -32, and (12, 5) collinear?
Make Sense? In Exercises 71–74, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
Determinants are used to write an equation of a line passing
through two points. An equation of the line passing through the
distinct points 1x1 , y12 and 1x2 , y22 is given by
71. I’m solving a linear system using a determinant that contains
two rows and three columns.
56. Are the points 1 -4, - 62, (1, 0), and (11, 12) collinear?
x
3 x1
x2
y
y1
y2
1
1 3 = 0.
1
Use this information to work Exercises 57–58.
57. Use the determinant to write an equation of the line passing
through 13, -52 and 1- 2, 62. Then expand the determinant,
expressing the line’s equation in slope-intercept form.
58. Use the determinant to write an equation of the line passing
through 1- 1, 32 and (2, 4). Then expand the determinant,
expressing the line’s equation in slope-intercept form.
Writing in Mathematics
59. Explain how to evaluate a second-order determinant.
60. Describe the determinants Dx and Dy in terms of the coefficients and constants in a system of two equations in two
variables.
61. Explain how to evaluate a third-order determinant.
62. When expanding a determinant by minors, when is it
necessary to supply minus signs?
63. Without going into too much detail, describe how to solve a
linear system in three variables using Cramer’s rule.
64. In applying Cramer’s rule, what does it mean if D = 0?
65. The process of solving a linear system in three variables
using Cramer’s rule can involve tedious computation. Is
there a way of speeding up this process, perhaps using
Cramer’s rule to find the value for only one of the variables? Describe how this process might work, presenting a
specific example with your description. Remember that
your goal is still to find the value for each variable in the
system.
66. If you could use only one method to solve linear systems in
three variables, which method would you select? Explain
why this is so.
Technology Exercises
67. Use the feature of your graphing utility that evaluates the
determinant of a square matrix to verify any five of the
determinants that you evaluated by hand in Exercises 1–10,
27–32, or 41–44.
72. I can speed up the tedious computations required by Cramer’s
rule by using the value of D to determine the value of Dx .
73. When using Cramer’s rule to solve a linear system, the number of determinants that I set up and evaluate is the same as
the number of variables in the system.
74. Using Cramer’s rule to solve a linear system, I found the
value of D to be zero, so the system is inconsistent.
75. a. Evaluate:
`
a
0
a
`.
a
a
3
b. Evaluate: 0
0
a
a
0
a
a3.
a
a
0
c. Evaluate: 4
0
0
a
a
0
0
a
a
a
0
a
a
4.
a
a
d. Describe the pattern in the given determinants.
e. Describe the pattern in the evaluations.
2
0
5
76. Evaluate: 0
0
0
0
3
0
0
0
0
0
2
0
0
0
0
0
1
0
0
0
05.
0
4
77. What happens to the value of a second-order determinant if
the two columns are interchanged?
78. Consider the system
a 1x + b 1y = c1
a 2x + b 2y = c2 .
Use Cramer’s rule to prove that if the first equation of the
system is replaced by the sum of the two equations, the
resulting system has the same solution as the original
system.
79. Show that the equation of a line through 1x1 , y12 and
1x2 , y22 is given by the determinant equation in Exercises
57–58.
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Page 868
868 Chapter 8 Matrices and Determinants
Group Exercise
81. Consider the equation
80. We have seen that determinants can be used to solve linear
equations, give areas of triangles in rectangular coordinates, and
determine equations of lines. Not impressed with these
applications? Members of the group should research an
application of determinants that they find intriguing. The group
should then present a seminar to the class about this application.
a. Set y = 0 and find the x-intercepts.
b. Set x = 0 and find the y-intercepts.
82. Divide both sides of 25x2 + 16y2 = 400 by 400 and simplify.
83. Complete the square and write the circle’s equation in
standard form:
Preview Exercises
x2 + y2 - 2x + 4y = 4.
Exercises 81–83 will help you prepare for the material covered in
the first section of the next chapter.
Chapter
y2
x2
+
= 1.
9
4
8
Then give the center and radius of the circle and graph the
equation.
Summary, Review, and Test
Summary
DEFINITIONS AND CONCEPTS
8.1
EXAMPLES
Matrix Solutions to Linear Systems
a. Matrix row operations are described in the box on page 807.
Ex. 1, p. 807
b. To solve a linear system using Gaussian elimination, begin with the system’s augmented matrix. Use matrix
row operations to get 1s down the main diagonal from upper left to lower right, and 0s below the 1s. Such a
matrix is in row-echelon form. Details are in the box on page 808.
Ex. 2, p. 809;
Ex. 3, p. 811
c. To solve a linear system using Gauss-Jordan elimination, use the procedure of Gaussian elimination, but
obtain 0s above and below the 1s in the main diagonal from upper left to lower right. Such a matrix is in
reduced row-echelon form. Details are in the box on page 813.
Ex. 4, p. 814
8.2
Inconsistent and Dependent Systems and Their Applications
a. If Gaussian elimination results in a matrix with a row containing all 0s to the left of the vertical line and a
nonzero number to the right, the system has no solution (is inconsistent).
Ex. 1, p. 818
b. In a square system, if Gaussian elimination results in a matrix with a row with all 0s, but not a row like the one
in part (a), the system has an infinite number of solutions (contains dependent equations).
Ex. 2, p. 820
c. In nonsquare systems, the number of variables differs from the number of equations.
Ex. 3, p. 821
8.3
Matrix Operations and Their Applications
a. A matrix of order m * n has m rows and n columns. Two matrices are equal if and only if they have the same
order and corresponding elements are equal.
Ex. 1, p. 827
b. Matrix Addition and Subtraction: Matrices of the same order are added or subtracted by adding or subtracting
corresponding elements. Properties of matrix addition are given in the box on page 830.
Ex. 2, p. 829
c. Scalar Multiplication: If A is a matrix and c is a scalar, then cA is the matrix formed by multiplying each element
in A by c. Properties of scalar multiplication are given in the box on page 831.
Ex. 3, p. 830;
Ex. 4, p. 831
d. Matrix Multiplication: The product of an m * n matrix A and an n * p matrix B is an m * p matrix AB. The
element in the ith row and jth column of AB is found by multiplying each element in the ith row of A by the
corresponding element in the jth column of B and adding the products. Matrix multiplication is not commutative:
AB Z BA. Properties of matrix multiplication are given in the box on page 836.
Ex. 5, p. 832;
Ex. 6, p. 833;
Ex. 7, p. 835
8.4
Multiplicative Inverses of Matrices and Matrix Equations
a. The multiplicative identity matrix In is an n * n matrix with 1s down the main diagonal from upper left to
lower right and 0s elsewhere.
b. Let A be an n * n square matrix. If there is a square matrix A-1 such that AA-1 = In and A-1 A = In , then
A-1 is the multiplicative inverse of A.
Ex. 1, p. 843
c. If a square matrix has a multiplicative inverse, it is invertible or nonsingular. Methods for finding multiplicative
inverses for invertible matrices, including a formula for 2 * 2 matrices, are given in the box on page 849.
Ex. 2, p. 844;
Ex. 3, p. 846;
Ex. 4, p. 848