1 Matrices: Definitions and Basic Operations
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16.1 Matrices: Definitions and Basic Operations
F1
E X A M P L E 3 Matrix equation
F2
The forces acting on a bolt are in equilibrium, as shown in Fig. 16.1. Analysing the
horizontal and vertical components as in Section 9.4, we find the following matrix
equation. Find forces F1 and F2.
28.1°
12.7°
437
8.0 N
c
3.5 N
Fig. 16.1
0.98F1  0.88F2
8.0
d = c
d
0.22F1 + 0.47F2
3.5
From the equality of matrices, we know that 0.98F1  0.88F2 = 8.0 and
0.22F1 + 0.47F2 = 3.5. Therefore, to find the forces F1 and F2, we must solve the system of equations
0.98F1  0.88F2 = 8.0
0.22F1 + 0.47F2 = 3.5
Using determinants, we have
F1 =
2 8.0
3.5
0.88 2
0.47
2 0.98
0.22
0.88 2
0.47
=
8.010.472  3.51 0.882
= 10.5 N
0.9810.472  0.221 0.882
Using determinants again, or by substituting this value into either equation, we find that
F2 = 2.6 N. These values check when substituted into the original matrix equation. ■
MATRIX ADDITION AND SUBTRACTION
If two matrices have the same number of rows and the same number of columns, their
sum is defined as the matrix consisting of the sums of the corresponding elements. If
the number of rows or the number of columns of the two matrices is not equal, they
cannot be added.
E X A M P L E 4 Adding matrices
(a)
J
8
0
1
2
5 9
3
R + J
3 7
6
4 6 0
8 + 1 32
R = J
2 6 5
0 + 6
= J
Practice Exercise
1. For matrices A and B, find A + B.
2 4
9 6
d B = c
d
A = c
7
5
7
3
5
6
1 + 4
2 + 1 22
5 + 6 9 + 0
R
3 + 6 7 + 5
5 1 9
R
4 9 12
(b) The matrices
3
C2
4
5 8
9 0S
2 3
and
3
C2
4
5 8 0
9 0 0S
2 3 0
cannot be added since the second matrix has one more column than the first matrix.
This is true even though the extra column contains only zeros.
■
The product of a number and a matrix (known as scalar multiplication of a matrix)
is defined as the matrix whose elements are obtained by multiplying each element of
the given matrix by the given number. Thus, we obtain matrix kA by multiplying the
elements of matrix A by k. In this way, A + A and 2A are the same matrix.
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438
CHAPTER 16 Matrices; Systems of Linear Equations
E X A M P L E 5 Scalar multiplication
For the given matrix A, find 2A:
A = c
5 7
d
3 0
2A = J
21 52
2132
2172
10 14
R = c
d
2102
6 0
■
By combining the definitions for the addition of matrices and for the scalar multiplication of a matrix, we can define the subtraction of matrices. That is, the difference of
matrices A and B is given by A  B = A + 1 B2. Therefore, we change the sign of
each element of B, and proceed as in addition.
E X A M P L E 6 Subtracting matrices
c
7
9
4
2 6
7
d  c
d = c
3
8 5
9
4
2
d + c
3
8
6
9
d = c
5
1
10
d
2
■
The operations of addition, subtraction, and multiplication of a matrix by a number
are like those for real numbers. For these operations, the algebra of matrices is like the
algebra of real numbers. We see that the following laws hold for matrices:
Practice Exercise
A + B
A + 1B + C2
k1A + B2
A + O
2. For matrices A and B in Practice
Exercise 1, find A  2B.
=
=
=
=
B + A
1A + B2 + C
kA + kB
A
(16.1)
(16.2)
(16.3)
(16.4)
(commutative law)
(associative law)
Here, we have let O represent the zero matrix. We will find in the next section that not
all laws for matrix operations are like those for real numbers.
E XE R C I SES 1 6 .1
In Exercises 1 and 2, make the given changes in the indicated
examples of this section and then perform the indicated operations.
1. In Example 4(a), interchange the second and third columns of the
second matrix and then add the matrices.
2. In Example 5, find the matrix  2A.
In Exercises 3–10, determine the value of the literal numbers in each
of the given matrix equalities.
3. c
5. c
a b
1
d = c
c d
4
x
r>4
2y
s
3
d
7
2
z
d = c
 5t
12
10
4
4. c
x
2
d = c d
x + y
5
9
d
5
6. 3a + bj 2c  dj 3e + f j4 = 35j a + 6 3b + c4
1j = 1 12
C + D
5
7. C 2C  D S = C 4 S
D  2E
6
8. c
2x  3y
13
d = c d
x + 4y
1
x  3 x + y
5 3
x
2
9. C x  z y + z S = c
d 10. c
d = c
4 1
x + y
4
x + t y  t
0
d
3
In Exercises 11–14, find the indicated sums of matrices.
11. c
12. c
2 3
1
d + c
5 4
5
1
3
0
5
50
13. £ 34
15
4.7
14. C 6.8
 1.9
7
d
2
9
4
d + c
2
2
1
0
7
d
3
82
 55
82
57 § + £ 45
14 §
62
26 67
2.1 9.6
4.9
4.8
7.4 S + C 3.4
0.7
5.9
5.6
9.6
0.7
10.1
 2.1
0.0 S
 1.6
In Exercises 15–34, use matrices A, B, and C to find the indicated
matrices.
A = c
1
2
4
6
7
d
11
B = c
7
4
9
1
6
d
8
C = c
3
7
19
d
5
15. A + B
16. A  B
17. A + C
19. 2A + B
20. 2B + A
21. A  2B
18. B + C
22. 3A  B
23. 4A
24. 3B
25. C  A
26.  12A + B
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16.2 Multiplication of Matrices
27. 3A
28.  2C
31. A  C
32.  4A  3B 33.  6B + A
29. A + 3B
30. B  0.5A
twocar garage. The following matrix shows the number of
houses of each type and the type of garage.
34. A  B + 2C
In Exercise 35–38, use matrices A and B to show that the indicated laws
hold for these matrices. In Exercise 35, explain the meaning of the result.
1
A = C 0
9
2
3
1
3
1
0
7
4S
2
1
0
11
4
B = C5
1
3
1
8
36. A + O = A
37.  1A  B 2 = B  A
38. 31A + B 2 = 3A + 3B
In Exercises 39 and 40, find the unknown quantities in the given
matrix equations.
39. An airplane is flying in a direction 21.0° north of east at 235 km>h
but is headed 14.5° north of east. The wind is from the southeast.
Find the speed of the wind vw and the speed of the plane vp relative to the wind from the given matrix equation. See Fig. 16.2.
c
Carport
1car garage
2car garage
0
1S
2
35. A + B = B + A
vp cos 14.5°  vw cos 45.0°
235 cos 21.0°
d = c
d
vp sin 14.5° + vw sin 45.0°
235 sin 21.0°
21.0
14.5
45.0
E
Fig. 16.2
40. Find the electric currents shown in Fig. 16.3 by solving the following matrix equation:
I1 + I2 + I3
0
C  2I1 + 3I2 S = C 24 S
 3I2 + 6I3
0
24 V
Type B
Type C
Type D
96
£ 62
0
75
44
35
0
24
68
0
0§
78
42. The inventory of a drug supply company shows that the following
numbers of cases of bottles of vitamins C and B3 (niacin) are in
stock: vitamin C—25 cases of 100mg bottles, 10 cases of 250mg bottles, and 32 cases of 500mg bottles; vitamin B3—30 cases
of 100mg bottles, 18 cases of 250mg bottles, and 40 cases of
500mg bottles. This is represented by matrix A below. After two
shipments are sent out, each of which can be represented by
matrix B below, find the matrix that represents the remaining
inventory.
A = c
vp
Type A
If the contractor builds two additional identical developments,
find the matrix showing the total number of each housegarage
type built in the three developments.
235 km/h
vw
439
25 10 32
d
30 18 40
B = c
10 5 6
d
12 4 8
43. One serving of brand K of breakfast cereal provides the given percentages of the given vitamins and minerals: vitamin A, 15%;
vitamin C, 25%; calcium, 10%; iron, 25%. One serving of brand G
provides: vitamin A, 10%; vitamin C, 10%; calcium, 10%; iron,
45%. One serving of tomato juice provides: vitamin A, 15%; vitamin C, 30%; calcium, 3%; iron, 3%. One serving of orangepineapple juice provides vitamin A, 0%; vitamin C, 100%; calcium,
2%; iron, 2%. Set up a tworow, fourcolumn matrix B to represent
the data for the cereals and a similar matrix J for the juices.
44. Referring to Exercise 43, find the matrix B + J and explain the
meaning of its elements.
2⍀
3⍀
6⍀
I1
I2
I3
Answers to Practice Exercises
Fig. 16.3
In Exercises 41–44, perform the indicated matrix operations.
1. c
7
0
 10
d
8
2. c
20
 21
8
d
1
41. The contractor of a housing development constructs four different
types of houses, with either a carport, a onecar garage, or a
16.2 Multiplication of Matrices
.VMUJQMJDBUJPOPG.BUSJDFT t *EFOUJUZ
.BUSJY t *OWFSTFPGB.BUSJY
The definition for the multiplication of matrices does not have an intuitive basis.
However, through the solution of a system of linear equations we can, at least in part,
show why multiplication is defined as it is. Consider Example 1.
E X A M P L E 1 Reasoning for the definition of multiplication
If we solve the system of equations
2x + y = 1
7x + 3y = 5
we get x = 2, y = 3. Checking this solution in each of the equations, we get
2122 + 11 32 = 1
7122 + 31 32 = 5
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440
CHAPTER 16 Matrices; Systems of Linear Equations
2 1
Let us represent the coefficients of the equations by the matrix c
d and the solu7 3
2
tions by the matrix c
d . If we now indicate the multiplications of these matrices and
3
perform it as shown
c
2 1
2
2122 + 11 32
1
dc
d = c
d = c d
7 3 3
7122 + 31 32
5
we note that we obtain a matrix that properly represents the rightside values of the
equations. (Note the products and sums in the resulting matrix.)
■
Following reasons along the lines indicated in Example 1, we now define the multiplication of matrices. If the number of columns in a first matrix equals the number of
rows in a second matrix, the product of these matrices is formed as follows: The element in a specified row and a specified column of the product matrix is the sum of the
products formed by multiplying each element in the specified row of the first matrix by
the corresponding element in the specific column of the second matrix. The product
matrix will have the same number of rows as the first matrix and the same number of
columns as the second matrix. We summarize this as follows.
Multiplication of Matrices
r If A is an n * k matrix, and B is a k * m matrix, then their product AB is an
n * m matrix.
r The ijth element of AB is formed by multiplying each element of row i in A
by the corresponding element of column j in B and then adding these
products.
r The product BA requires that n = m, so just because you can form the product AB does not mean that you can form the product BA. Clearly, AB ≠ BA
in general, so that matrix multiplication is not commutative.
E X A M P L E 2 Multiplying matrices
Find the product AB and the product BA, where
2 1
A = C 3 0 S
1 2
B = c
1 6 5
3 0 1
2
d
4
Matrix A is 3 * 2, and matrix B is 2 * 4, so the product AB can be formed, resulting in
a 3 * 4 matrix. The calculations are shown below, with the elements used to form the
element AB11 and the element AB32 outlined in colour.
2 1
1 6 5
C 3 0 S J
3 0 1
1 2
21 12 + 1132
2
R = C 31 12 + 0132
4
11 12 + 2132
1
= C3
5
Practice Exercise
1. Find the product AB.
1
4
5
A = C 8 2 S B = c
7
0 12
3
d
10
12
18
6
11
15
7
2162 + 1102
3162 + 0102
1162 + 2102
2152 + 1112
3152 + 0112
1152 + 2112
21 22 + 11 42
31 22 + 01 42 S
11 22 + 21 42
8
6S
10
In trying to form the product BA, we see that B has four columns and A has three
rows. Since these numbers are not the same, the product cannot be formed.
■
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16.2 Multiplication of Matrices
441
E X A M P L E 3 Multiplying matrices
■ Note that, if the second matrix had been to
the left of the first, then the product would have
had four rows and four columns.
c
1 9
2 0
3
7
6
2 1
dD
1 3
3
The product of two matrices below may be formed because the first matrix has four
columns and the second matrix has four rows. The matrix is formed as shown.
2
0
1162 + 9112 + 3132 + 1 22 132
T = c
5
2162 + 0112 + 1 72 132 + 1132
9
6 + 9 + 9  6
12 + 0  21 + 3
6 31
= c
d
6
40
= c
2 + 0  15  18
d
4 + 0 + 35 + 9
11 22 + 9102 + 31 52 + 1 22 192
d
21 22 + 0102 + 1 72 1 52 + 1192
■
IDENTITY MATRIX
There are two special matrices of particular importance in the multiplication of matrices. The first of these is the identity matrix I, which is a square matrix with 1’s for
elements of the principal diagonal with all other elements zero. (The principal diagonal
starts with the element a11.) It has the property that if it is multiplied by another square
matrix with the same number of rows and columns, then the second matrix equals the
product matrix.
E X A M P L E 4 Identity matrix
Show that AI = IA = A for the matrix
A = c
2
4
3
d
1
Since A is 2 * 2, we choose I to be 2 * 2. Therefore, for this case,
I = c
1 0
d
0 1
elements of principal diagonal are 1’s
Forming the indicated products, we have results as follows:
2 3 1 0
dc
d
4
1 0 1
2112 + 1 32 102 2102 + 1 32 112
2
= c
d = c
4112 + 1102
4102 + 1112
4
1 0 2 3
IA = c
dc
d
0 1 4
1
1122 + 0142 11 32 + 0112
2 3
= c
d = c
d
0122 + 1142 01 32 + 1112
4
1
AI = c
Therefore, we see that AI = IA = A.
3
d
1
■
INVERSE OF A MATRIX
For a given square matrix A, its inverse A1 is the other important special matrix. The
matrix A and its inverse A1 have the property that
AA1 = A1A = I
(16.5)
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442
CHAPTER 16 Matrices; Systems of Linear Equations
If the product of two square matrices equals the identity matrix, the matrices are called
inverses of each other. Under certain conditions, the inverse of a given square matrix
may not exist. In the next section, we develop the procedure for finding the inverse of a
square matrix, and the section that follows shows how the inverse is used in the solution of systems of equations. At this point, we simply show that the product of certain
matrices equals the identity matrix and that therefore these matrices are inverses of
each other.
E X A M P L E 5 Inverse matrix
For the given matrices A and B, show that AB = BA = I, and therefore that B = A1:
A = c
Practice Exercise
2. Show that AB = BA = I.
5 7
3
A = c
d B = c
2
3
2
1
2
3
d
7
B = c
7 3
d
2 1
Forming the products AB and BA, we have the following:
7
d
5
1 3 7
dc
2
7 2
7 3
1
BA = c
dc
2 1 2
AB = c
3
7  6
3  3
1 0
d = c
d = c
d
1
14 + 14 6 + 7
0 1
3
7  6 21 + 21
1 0
d = c
d = c
d
7
2  2
6 + 7
0 1
Since AB = I and BA = I, B = A1 and A = B1.
■
E X A M P L E 6 Matrix multiplication—application
A company makes three types of automobile parts. In one day, it produces 40 of type
X, 50 of type Y, and 80 of type Z. Required for production are 4 units of material and 1
workerhour for type X, 5 units of material and 2 workerhours for type Y, and 3 units
of material and 2 workerhours for type Z. By representing the number of each type
produced as matrix A and the material and time requirements as matrix B, we have
units of workerhours
material
type X type Y
A = 3 40
50
type Z
804
number of each type
produced
4 1
B = £5 2§
3 2
type X
type Y
type Z
material and time
required for each
The product AB gives the total number of units of material and the total number of
workerhours needed for the day’s production in a onerow, twocolumn matrix:
4 1
AB = 3 40 50 80 4 £ 5 2 §
3 2
total
units of
material
total
workerhours
= 3 160 + 250 + 240 40 + 100 + 1604 = 3 650 3004
Therefore, 650 units of material and 300 workerhours are required.
y
■
E X A M P L E 7 Multiplication of matrices—application to robotic arms
2m
45∘
45∘
Fig. 16.4
x
Consider the twodimensional robotic arm in Fig. 16.4, consisting of two links, each one
2.0 m long. The first link has been rotated 45° with respect to the xaxis, and the second
link has been rotated 45° with respect to the first link. We can find the coordinates of the
end of the arm by applying transformation matrices to the origin of our coordinate system.
x
We start by representing a point (x, y) as the vector (a 3 * 1 matrix) £ y § . In par0
1
ticular, the origin is represented by the vector £ 0 § . If the coordinate system is rotated
1
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443
16.2 Multiplication of Matrices
by an angle u and then translated by m units in the xdirection and n units in the ydirection, the new coordinates of the origin are found by the matrix multiplication
■ See the chapter introduction.
sin u
cos u
0
cos u
£ sin u
0
m 0
n § £0§
1 1
For our example, u = 45°, m = 2 (the length of a link), and n = 0, so the transformation matrix becomes
cos 45°
£ sin 45°
0
sin 45°
cos 45°
0
1
2
12
1
0 § = £ 12
1
0
1
12

1
12
0
2
0§
1
Since the process is repeated twice (once for each link), the location of the end of the
arm is given by
£
Some Properties of Matrix Algebra
1
12
1
12
0
1
 12
1
12
0
1
2 12
1
0 § £ 12
1
0
1
 12
1
12
0
2 0
0
0§ £0§ = £1
1 1
0
1
0
0
2 + 22 0
2 + 22
22 § £ 0 § = £ 22 §
1
1
1
which represents the point 12 + 22,222.
This method can be generalized to robotic arms in three dimensions by including
more variables. Moreover, transformation matrices are also useful for computer graph■
ics and CAD.
We have seen that matrix multiplication is not commutative (see Example 2); that is,
AB ≠ BA in general. This differs from the multiplication of real numbers. Another
difference is that it is possible that AB = O, even though neither A nor B is O.
Moreover, the only number that does not have a multiplicative inverse is 0, yet there
are many nonzero matrices that do not have an inverse. There are also some similarities, for example, AI = A, where I and the number 1 are equivalent. Also, the distributive property A1B + C2 = AB + AC holds for matrix multiplication.
E XE RC IS ES 1 6 .2
In Exercises 1 and 2, make the given changes in the indicated examples
of this section and then perform the indicated multiplications.
1. In Example 2, interchange columns 1 and 2 in matrix A and then
do the multiplication.
2. In Example 5, in A change  2 to 2 and 3 to 3, in B change 2 to
 2 and 3 to  3, and then do the multiplications.
In Exercises 3–14, perform the indicated multiplications.
3. 34
 24 c
2
5. c
0
3
7
8
7. £
7
2
6
3
4
1 0
d
2 6
90
1
d £  25 §
3
50
1
8 § c 4
2
4
5
3
d
5
4. 3  12
0
6. c
4
12
8. £ 43
36
6
4
 23 4 £ 13
0
3
1 2
d £1
11 2
6
47
25
18 § c d
66
22
8
 12 §
9
1
2§
1
1
3
9. D
10
5
11. c
13. c
14. c
2
5
7
5 2
Tc
1 5
12
3 3
dc
1 7
9.2
3.8
1 2
2 4
0
5
1
d
3
1
d
8
6.5
2.3 0.5
d £ 4.9
2.4 9.2
1.8
1
1
6 6 1
d E 0U
0 1 2
5
2
10. 35 44 c
12. c
5.2
1.7 §
6.9
7
5
4
5
4
d
5
8 90
dc
0
10
100
d
40
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444
CHAPTER 16 Matrices; Systems of Linear Equations
In Exercises 15–18, find, if possible, AB and BA. If it is not possible,
explain why.
15. A = 31
16. A = c
17. A = c
3
1
2
B = £ 4
5
2 0
d
4 5
 10
42
2
18. A = £ 3
0
1
B = £ 5§
7
 3 84
1
1
2
7
0§
1
B = 34
1
21. A = £ 2
1
3
0
2
0
6 §
1
5
2
3
24. A = c
5
1
25. A = £ 2
1
1
26. A = £ 3
2
2
d
1
4
d
7
2
5
3
1
4
3
7
B = c
5
3
7§
5
3
8§
4
4
B = £3
1
8
B = £ 4
1
39. Show that A2  I = 1A + I 2 1A  I 2 for A = c
28
d
64
2 0
3 1 §
1 3
1
2
1
5
2
1
1
1 §
1
4
1 §
1
In Exercises 27–30, determine by matrix multiplication whether or not
A is the proper matrix of solution values.
1
27. 3x  2y =  1
A = c d
2
4x + y = 6
28. 4x + y =  5
3x + 4y = 6
A = c
29. 3x + y + 2z = 1
x  3y + 4z =  3
2x + 2y + z = 1
30. 2x  y + z = 7
x  3y + 2z = 6
3x + y  z = 8
2
2 § , show that A2  4A  5I = O.
1
j 0
d , where j = 1 1, show that J 2 =  I, J 3 =  J,
0 j
and J 4 = I. Explain the similarity with j 2, j 3, and j 4.
1 2
d
2 5
4
d
2
2
1
2
38. For J = c
 1 54
1
22. A = £ 4
2
B = c
a b
c d
d and B = c
d , show that AB = BA.
b a
d c
37. Using two rows and columns, show that 1  I 2 2 = I.
In Exercises 23–26, determine whether or not B = A1.
23. A = c
34. Show that A3B3 = AB
1
36. For matrix A = £ 2
2
15
20. A = c
5
5
1§
4
33. Show that B3 = B
35. For matrices A = c
In Exercises 19–22, show that AI = IA = A.
1 8
19. A = c
d
2 2
32. Show that C 2 = O.
In Exercises 35–46, perform the indicated matrix multiplications.
6
B = £  15 §
12
25 40
d
5 0
31. Show that A2 = A.
2
d
3
1
A = £ 2§
1
3
A = £ 2 §
1
In Exercises 31–34, perform the indicated matrix multiplications,
using the following matrices. For matrix A, A2 = A * A.
2 3 5
1  2 6
1 3  4
4
5 § B = £ 3
2
9 § C = £ 1
3
4§
A = £ 1
1 3 4
2
0 3
1 3 4
2 4
d.
3 5
40. In the study of polarized light, the matrix product
1 0
1 0
1
c
d c
d c d occurs 1j = 1 12 . Find this product.
0 j 1 j 1
41. In studying the motion of electrons, one of the Pauli spin matrices
0 j
d , where j = 1 1. Show that s2y = I.
used is sy = c
j 0
42. In analysing the motion of a robotic mechanism, the following
matrix multiplication is used. Perform the multiplication and
evaluate each element of the result. (See Example 7.)
sin 60°
cos 60°
0
cos 60°
£ sin 60°
0
0 2
0§ £4§
1 1
43. In an ammeter, nearly all the electric current flows through a
shunt, and the remaining known fraction of current is measured
by the metre. See Fig. 16.5. From the given matrix equation, find
voltage V2 and current i2 in terms of V1, i1, and resistance R,
whichever may be applicable.
c
V2
d = £
i2
1
1
R
0
1
§c
V1
d
i1
V2
i1
i2
Shunt
R
V1
Fig. 16.5
44. In the theory related to the reproduction of colour photography,
the equation
X
1.0 0.1 0
x
£ Y § = £ 0.5 1.0 0.1 § £ y §
Z
0.3 0.4 1.0
z
is found. The X, Y, and Z represent the red, green, and blue densities of the reproductions, respectively, and the x, y, and z represent the red, green, and blue densities, respectively, of the subject.
Give the equations relating X, Y, and Z and x, y, and z.
45. The path of an earth satellite can be written as
3x y4 c
7.10
1 x
d c d = 35.13 * 108 4
1 7.23 y
where distances are in kilometres. What type of curve is represented? (See Section 14.1.)
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16.3 Finding the Inverse of a Matrix
46. Using Kirchhoff’s laws on the circuit shown in Fig. 16.6, the following matrix equation is found. By matrix multiplication, find
the resulting system of equations.
£
R1 + R2
 R2
0
 R2
R2 + R3 + R4
 R4
R1
V1
0
I1
V1
 R4 § £ I2 § = £ 0 §
R4 + R5
I3
 V2
R3
R2
I1
445
Answers to Practice Exercises
43
 33
1. £ 54 44 §
 84 120
2. AB = BA = c
10
d
01
R5
R4
I2
V2
I3
Fig. 16.6
16.3 Finding the Inverse of a Matrix
Inverse of a 2 : 2.BUSJY t (BVTTo+PSEBO
.FUIPE t *OWFSTFPOB$BMDVMBUPS
In this section we discuss two methods for finding the inverse of a square matrix. The
first method is straightforward, but it can only be used for 2 * 2 matrices.
The Inverse of a 2 : 2 Matrix
1. Evaluate the determinant of the matrix. If the determinant is zero, the inverse
does not exist.
2. Interchange the elements on the principal diagonal.
3. Change the signs of the offdiagonal elements.
4. Divide each resulting element by the determinant.
This method is illustrated in the following example.
E X A M P L E 1 Inverse of a 2 : 2 matrix—method 1
2 3
d.
4 7
First, we find the determinant of the original matrix, which means we evaluate
Find the inverse of the matrix A = c
22
4
3 2
= 14  1 122 = 2
7
Since the determinant is not zero, the inverse exists. Next, we interchange the elements
on the principal diagonal and change the signs of the offdiagonal elements. This gives
us the matrix
c
7 3
d
4 2
signs changed
elements interchanged
We now divide each element of the second matrix by 2 (the determinant). This gives
Practice Exercise
1. Find the inverse:
3 8
A = c
d
1 2
■ See Exercise 37.
A1
1 7
=
c
2 4
7
3
2
d =
2
4
2
3
7 3
2
Ơ = Ê2 2Đ
2
2 1
2
inverse
Checking by multiplication gives
AA1 = c
2
4
3 72
dc
7 2
 32
7  6
d = c
1
14  14
3 + 3
1 0
d = c
d = I
6 + 7
0 1
Since AA1 = I, the matrix A1 is the proper inverse matrix.
■
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446
CHAPTER 16 Matrices; Systems of Linear Equations
■ The Gauss–Jordan method is named for the
German mathematician Karl Gauss (1777–1855)
and the German geodesist Wilhelm Jordan
(1842–1899).
The second method, called the Gauss–Jordan method, is applicable for square
matrices of any size.
5IF*OWFSTFPGB.BUSJYCZUIF(BVTTo+PSEBO.FUIPE
1. Set up the given matrix and the identity matrix of the same size side by side.
2. Transform the given matrix into the identity matrix by performing any of the
following allowable row operations:
a. Any two rows may be interchanged.
b. Every element in any row may be multiplied by any number other than
zero.
c. Any row may be replaced by a row whose elements are the sum of a
nonzero multiple of itself and a nonzero multiple of another row.
Work one column at a time, transforming the columns in order from left to
right.
3. At every step, perform the same row operations on the identity matrix. The
resulting matrix will be the required inverse.
Note that these are row operations, not column operations, and that they are the operations used in solving a system of equations by addition and subtraction.
E X A M P L E 2 2 : 2*OWFSTF(BVTTo+PSEBONFUIPE
Practice Exercise
2. Find the inverse using the Gauss–Jordan
3 8
method: A = c
d
1 2
Find the inverse of the matrix
A = c
2
4
3
d
7
this is the same matrix as in Example 1
First, we set up the given matrix with the identity matrix side by side.
c
2
4
3 1 0
`
d
7 0 1
The vertical line simply shows the separation of the two matrices.
We wish to transform the left matrix into the identity matrix. Therefore, the first
requirement is a 1 for element a11. Therefore, we divide all elements of the first row
by 2. This gives the following setup:
LEARNING TIP
As in Example 2, (1) always work one
column at a time, from left to right,
and (2) never undo the work in a previously completed column.
c
1
4
 32 12 0
`
d
7 0 1
Next, we want to have a zero for element a21. Therefore, we subtract 4 times each element of row 1 from the corresponding element in row 2, replacing the elements of
row 2. This gives us the following setup:
c
1
4  4112
1
 32
2
`
7  41  32 2 0  41 12 2
0
d
1  4102
or
c
1
0
1
 32
0
` 2
d
1 2 1
Next, we want to have 1, not 1, for element a22. Therefore, we multiply each element of row 2 by 1. This gives
c
1
0
 32 12
`
1 2
0
d
1
Finally, we want zero for element a12. Therefore, we add 32 times each element of row 2
to the corresponding elements of row 1, replacing row 1. This gives
c
1 + 32 102
0
 32 + 32 112
`
1
1
2
+ 32 122
2
0 + 32 1 12
d
1
or
c
1 0 72
`
0 1 2
 32
d
1
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447
16.3 Finding the Inverse of a Matrix
At this point, we have transformed the given matrix into the identity matrix, and the
identity matrix into the inverse. Therefore, the matrix to the right of the vertical bar in
the last setup is the required inverse. Thus,
 32
d
1
7
A1 = c 2
2
■
As was done in Example 2, it is generally best to make the element on the principal
diagonal for the column 1 first and then make all other elements in the column 0. Let us
consider two more examples.
E X A M P L E 3 2 : 2*OWFSTF(BVTTo+PSEBONFUIPE
Find the inverse of the matrix c
original setup
c
3 6 2 1 0
d
4 5 0 1
3 6
d.
4 5
c
row 1 divided by  3
c
1
4
2
`
5
 4 times row 1
added to row 2
Therefore, A1 = c
5
 39
4
39
2
13
1 d,
13
2 2  13
4
13
3
0
d
1
row 2 divided by 13
 13
0
d
0 1
1
0
c
1
0
2
`
1
 13
4
39
c
5
1 0 2  39
4
0 1
39
A1
I
2
13
1 d
13
0
1 d
13
2 times row 2
added to row 1
which can be checked by multiplication.
■
E X A M P L E 4 3 : 3*OWFSTF(BVTTo+PSEBONFUIPE
1
Find the inverse of the matrix £ 3
2
original setup
1
£ 3
2
2
5
1
1
£ 0
2
2
1
1
1 1 0 0
1 3 0 1 0 §
2 0 0 1
1 2
£0 1
0 3
1 1
2 3 3
4 2
1
1 0 0
3
2 3 1 0 §
2
0 0 1
1 0
£0 1
0 3
3 5
2 3 3
4
2
1 0
£0 1
0 0
3 5
2 3 3
2 7
 3 times row 1
added to row 2
2 times row 1
added to row 3
1
£0
0
2
1
3
1
1 0 0
2 3 3 1 0 §
4
2 0 1
row 2 multiplied by  1
1
1 § .
2
2
5
1
 2 times row 2
added to row 1
0 0
1 0 §
0 1
 3 times row 2
added to row 3
row 3 divided by 2
1 0
£0 1
0 0
3 5
2 3 3
1  72
2 0
1 0 §
3
2
2 times row 3
added to row 2
1
2
2 0
1 0 §
0 1
1 0 3 5 2 0
£ 0 1 0 3 4 2 0 §
0 0 1  72 32 12
2 0
1 0 §
3 1
1 0 0 11
2
£ 0 1 0 3 4
0 0 1  72
 3 times row 3
added to row 1
I
 52
2
3
2
A1
 32
1§
1
2
Therefore, the required inverse matrix is
11
2
£ 4
 72
which may be checked by multiplication.
 52
2
3
2
 32
1§
1
2
■