syllabus
syllabus
rref
efer
erence
ence
Core topic:
Matrices and applications
In this
In this
cha
chapter
pter
3A Operations with matrices
3B
Multiplying matrices
3C Powers of a matrix
3D Multiplicative inverse and
solving matrix equations
3E
The transpose of a matrix
3F
Applications of matrices
3G Dominance matrices
3
Matrices
Introduction to matrices
Four towns are connected by roads as shown in the figure. There is one road connecting A and B, two roads connecting A and C and so on. This information may be repre-sented as shown in the table.
If the headings at the top and side of this display are removed, an array of numbers only is left:
This array of numbers is called a matrix (plural, matrices).
The arrangement of numbers in matrices is an extension of our number system and, as we will see, the rules that govern matrix calculations have many similarities with the arithmetic of numbers. Matrices are particularly useful in solving complex problems in linear programming.
A matrix is a rectangular array of numbers arranged in rows and columns. The numbers in the matrix are called the elements of the matrix.
The matrix above is a 4 × 4 matrix as it has 4 rows and 4 columns. We say the order of the matrix is 4 by 4.
The matrix is a 3 × 2 matrix because it has 3 rows and 2 columns. Note the
square brackets used to enclose the array.
A matrix with m rows and n columns is called an m×n matrix. We say the order of the matrix is m×n. The dimensions of a matrix are always given as the number of rows multiplied by the number of columns.
The elements of the matrix are referred to by the row and then by the column pos-ition. In the 3 × 2 matrix above, the row 1, column 1 element is 2, the row 3, column 1 element is −1 and the row 1, column 2 element is 0.
To
A B C D
From
A 0 1 2 0
B 1 0 0 1
C 2 0 0 3
D 0 1 3 0
A
B
C
D
0 1 2 0
1 0 0 1
2 0 0 3
0 1 3 0
2 0 1 –4 1
C h a p t e r 3 M a t r i c e s
125
We often use capital letters as symbols for matrices. Thus we may write
A=
In general, the elements of a matrix A are referred to as ai j where i refers to the
row position and j refers to the column position.
That is, A = , depending on the order of the matrix
where A is an m×n matrix.
The row 1, column 1 element is a1 1.
The row 3, column 1 element is a3 1 and so on.
2 0
1 –4 1
– 2
a11 a12 a13 a14 … a1n
a21 a22 a23 a24 … a2n
a31 a32 a33 a34 … a3n
am1 am2 am3 am4 … amn
. . . . ... .
. . . . ... .
. . . . ... .
For each of the following give the order of the matrix, if it exists, and where possible write down the elements in row 2, column 1 and row 1, column 3.
A = B = C = D =
THINK WRITE
A has 3 rows of numbers and 2 columns of numbers.
A is a 3 × 2 matrix.
B has 2 rows and 3 columns. B is a 2 × 3 matrix.
C has 3 rows and 1 column. C is a 3 × 1 matrix.
D is not a rectangular array of numbers as it does not have all positions filled.
D is not a matrix.
The row 2, column 1 element is the number in the second row and the first column.
The row 1, column 3 element is the number in the first row and the third column.
In A and C there is no row 1, column 3 element since there is no third column in either matrix.
2 5 3 6 4 7
1 2 3 1 – –2 –3
1 –
2 –
3 –
5 0 2 6
1
2
3
4
5
Matrix
Row 2, column 1
element
Row 1, column 3
element
A 3 —
B –1 3
C –2 —
6
7
1
WORKED
Example
Operations with
matrices
Addition
The sports coordinator at Mathglen State High School kept records of the number of first, second and third ribbons awarded to competitors in each house at the swimming and athletics carnivals and sports events.
The results were:
To find the total number of first, second and third places for each house, the swimming, athletics and sports results may be added. The elements in corresponding positions are added to give the total number of first, second and third places for each house:
Adding the elements for each event results in the following matrix:
Addition of matrices is performed by adding corresponding elements.
Subtraction
The subtraction of matrices is also performed by the usual rules of arithmetic on corresponding elements of the matrices. It follows that:
House Swimming House Athletics and sports
1st 2nd 3rd 1st 2nd 3rd
Hamilton 25 28 24 Hamilton 35 35 27
Leslie 38 30 35 Leslie 33 34 39
Barnes 34 36 35 Barnes 30 33 36
Cunningham 35 38 38 Cunningham 34 34 30
House 1st 2nd 3rd
Hamilton Leslie Barnes Cunningham
25 + 35 = 60 38 + 33 = 71 34 + 30 = 64 35 + 34 = 69
28 + 35 = 63 30 + 34 = 64 36 + 33 = 69 38 + 34 = 72
24 + 27 = 51 35 + 39 = 74 35 + 36 = 71 38 + 30 = 68
C h a p t e r 3 M a t r i c e s
127
1. Subtraction of matrices is performed by subtracting corresponding elements. 2. Addition and subtraction of matrices can be performed only if the matrices are
of the same order; that is, they have the same number of rows and columns.
Furthermore, addition of matrices is commutative. That is, for two matrices A and B of the same order:
A+B=B+A
Multiplication by a scalar
Consider the matrix B=
To find 3B we could use repeated addition: 3B=B+B+B
=
=
3B could have been calculated more efficiently by multiplying each element of B by 3.
Thus 3B= 3 = =
If A= B= C=
find, if possible:
a A+B b A−B c B−C.
THINK WRITE
a Add the numbers in the corresponding positions of each matrix.
a A+B=
=
b Subtract the numbers in the corresponding positions of each matrix.
b A−B=
=
c Subtraction cannot be performed since the order of B is 2 × 2 and the order of C is 2 × 3.
c B−C cannot be calculated because B
and C are of different orders.
1 2 3 4
1 4 2 3
2 2 0 2 2 0
1 2 3 4
1 4 2 3 +
2 6 5 7
1 2 3 4
1 4 2 3 –
0 –2 1 1
2
WORKED
E
xample
1 4 2 3
1 4 2 3
1 4 2 3
1 4 2 3
+ +
3 12 6 9
1 4 2 3
3×1 3×4
3×2 3×3
3 12
6 9
The number 3 in the term 3B is called a scalar because it is a constant. Terms such as 3B refer to scalar multiplication of matrices.
When a matrix is multiplied by a scalar, each element of the matrix is multiplied by the scalar.
If and find:
a 2A b 5B c 2A+ 5B d 4(A+B) e 2(B−A).
THINK WRITE
a Multiply each element of A by 2. a 2Α =
=
b Multiply each element of B by 5. b 5Β =
=
c Add the two matrices found in parts a and b.
c 2Α + 5Β =
=
d Add A and B inside the brackets. d 4(Α + Β) =
= 4
Multiply each element of the resulting matrix by 4.
=
e Subtract A from B (inside the brackets). e 2(Β − Α) =
= 2
Multiply each element of the resulting matrix by 2.
=
A 2 –3
4 1
= B 3 3
3 – –2 =
2 2 –3 4 1 4 –6 8 2
5 3 3 3 – –2
15 15 15 – –10
4 –6 8 2
15 15 15 – –10 +
19 9 7 – –8
1 4 2 –3
4 1
3 3 3 – –2 +
5 0 1 –1
2 20 0
4 –4
1 2 3 3
3 – –2
2 –3 4 1 –
1 6 7 – –3
2 2 12
14 – –6
3
C h a p t e r 3 M a t r i c e s
129
There are some obvious but important features of scalar multiplication. If A and B are matrices of the same order and a, b are real numbers, then:
1. aA + bA = (a + b)A 2. aA + aB = a(A + B) 3. (ab)A = a(bA)
Operations 1 and 2 are similar to the Distributive Law of Addition. Operation 3 is similar to the Associative Law of Multiplication.
If aA = 0, then a = 0 or A is a zero matrix. A zero matrix is a matrix which has all elements equal to zero.
Equality of matrices
This leads to an important principle about the equality of matrices.
Two matrices are equal if they are of the same order and all corresponding
elements are equal; that is, if A = and B = then A = B.
Simple matrix equations
We know that to solve an algebraic equation such as 2x + 5 = 11, we: 1. subtract 5 from both sides to obtain 2x = 11 − 5 which gives 2x = 6
2. then, divide both sides by 2 (or multiply by ) to obtain x = 6 × or x = 3.
Simple matrix equations which require the addition or subtraction of a matrix or multiplication of a scalar can be solved in similar ways to those employed with algebraic equations.
a b c d
a b c d
1 2
--- 1
2
---Solve the following matrix equations.
a b c
Continued over page
THINK WRITE
a To get A by itself multiply both sides by .
a 5Α =
Α =
Simplify the matrix A. Α = ⇒
5A 50 35
15
– 20
= P 3 2
1 5
+ 6 9
2 – 4
= 2B 1 2 –3
2 0 1
– 3 4 7
2 – 6 –5 =
1
1 5
---50 35 15
– 20
1 5
--- 50 35
15
– 20
2 10 7
3
– 4
4
WORKED
E
xample
THINK WRITE
b To get P by itself subtract from
both sides.
b
Simplify the matrix P. ⇒
c First get 2B by itself by adding
to both sides.
c
Simplify the right-hand side.
Multiply both sides by to get B by itself. Β =
Simplify the matrix B. Β =
1 3 2
1 5 P
3 2 1 5
+ 6 9
2 – 4 =
P 6 9
2 – 4
= 3 2
1 5 –
2 P 3 7
3 – –1 =
1
1 2 –3 2 0 1
2B 1 2 –3
2 0 1
– 3 4 7
2 – 6 –5 =
2B 3 4 7
2 – 6 –5
1 2 –3 2 0 1 +
=
2 4 6 4
0 6 –4 =
3 12--- 1
2
--- 4 6 4
0 6 –4
4 2 3 2
0 3 –2
remember
1. A matrix is a rectangular array of numbers arranged in rows and columns. 2. An m×n matrix has m rows and n columns.
3. The numbers in the matrix are called the elements of the matrix.
4. Elements are referred to by the row and column position. For example, ai j refers to the ‘i th’ row and the ‘j th’ column of matrix A.
5. Addition of matrices is performed by adding corresponding elements.
6. Subtraction of matrices is performed by subtracting corresponding elements. 7. Addition and subtraction of matrices can be performed only if the matrices are
of the same order.
8. When a matrix is multiplied by a scalar, each element of the matrix is multiplied by that scalar.
9. Two matrices are equal if they are of the same order and all corresponding elements are equal.
C h a p t e r 3 M a t r i c e s
131
Operations with matrices
1 Using a table format, give the order of each of the following matrices and where possible write down the row 2, column 1 and row 1, column 3 elements of each.
, , , ,
2 If , and , calculate:
3 Using the matrices A, B and C from question 2 find:
4
Use , , , and
to answer questions a to e.
a The order of D is:
b Which one of the following cannot be calculated?
c 2A + 3E is equal to:
d 3C − 4B is equal to:
e The element e3 2 is equal to:
5 If and , calculate:
a A + B b A − B c B + C d C − A
a 2A b 2A − B c 2A + 3B d 3(A + B) e 2A + 3B − 4C
A 3 × 2 B 4 × 2 C 2 × 4 D 3 × 3 E 2 × 2
A A + E B B + C C 4D D A + B E 6(E − A)
A B C D E
A B C D E
A 9 B 4 C 3 D 7 E −1
a C + D b C − D c 2C d 2C + D e 2(C + D) f −D
3A
WORKED
Example
1
A –5 2 8 4
= B
6 5 7
= C = 1 8 10 20 D 4 4 4
4 4 4
= E
5 0 2 1 1 8 0 –5 3 =
Mathca
d Operations with matrices WORKED Example 2
A –2 3 4 7
= B 5 0
4 – 2
= C 1 –3
2 6 =
SkillS HEET
3.1
WORKEDExample
3
m
multiple choiceultiple choice
A
1 2 3 2 – –1 4
6 3 0
= B 5 –4
1 3
= C –4 3
2 7
= D 2 3 4 1
0 –2 7 5 =
E
0 5 –1 2 3 9 6 4 –2 =
3 16 7 2 – 3 30 30 17 –6
1 7 2 0 2 13 12 7 –2
4 15 –2 6 3 8 9 7 2
2 16 3 2 – 7 35 30 18 –6
2 19 3 2 7 35 30 18 –6
32 – 25
2 9
8 –7 10 33 31 – 0 5 19 1 0 0 1 32 – –7
5 9
C
1 4 7 2 5 8 3 6 9
= D
1 –4 7 2 –5 –8 3 –6 9 =
6 Solve the following matrix equations.
a b
c d
7 Explain why the following matrix equation has no solution.
8 Write down the matrix representing the following maps in the form:
Use alphabetic order for the sequencing of rows and columns.
9 A mathematically inclined student has decided to keep a record of her test results in matrix form. Her results so far are Maths B tests: 82%, 75% and 91%; Maths C tests: 54%, 68% and 82%.
Write these results in a 3 × 2 matrix.
10 Place the following sporting results in a suitable matrix format.
a Brisbane Lions 15 goals 14 behinds 104 points defeated Geelong 7 goals 10 behinds 52 points.
b Adelaide Crows have played 13 games for 7 wins, 5 draws and 1 loss; they have scored 31 goals for and 18 against; their points score is 26. Fremantle have played 12 games for 4–4–4; their goals are 17–15 and their points score is 16.
11 Adelaide Crows defeat Fremantle 4 goals to 1. Update the matrix in question 10b (note that 3 points are awarded for a win and 0 for a loss).
12 Write down any 2 × 2 matrices called A, B and C. Check if the following are true. a A + B = B + A
b (A + B) + C = A + (B + C) c A − B = B − A
d 2A + 2C = 2(A + C) WORKED
Example
4
3P 6 0
9 –3
= Q 4 0
1 4
+ 2 0
5 – 6 =
3M –2 0 3 4 6 –1
– –1 0 0
2 3 –2
= 26 ––26
3 –3 N –
0 4 7 6 5 –2 =
2A 4 8
4 – 0
+ 5 1 3
2 – 5 2 =
Number of
routes to Number
of routes from
b
a A B
C
D
J
L
C h a p t e r 3 M a t r i c e s
133
Multiplying matrices
The sports results at Mathglen State High School were:
To calculate the total points for each house, this matrix is multiplied by since 5 points are awarded for first, 3 for second and 1 for third.
The result can be obtained using the following operations. Hamilton: 60 × 5 + 63 × 3 + 51 × 1 = 540
Leslie: 71 × 5 + 64 × 3 + 74 × 1 = 621 Barnes: 64 × 5 + 69 × 3 + 71 × 1 = 598 Cunningham: 69 × 5 + 72 × 3 + 68 × 1 = 629
We can also write A × B = C, where , and
The order of A is 4 × 3, B is 3 × 1 and C is 4 × 1.
Therefore, a 4 × 3 matrix multiplied by a 3 × 1 matrix gives a 4 × 1 matrix. Two matrices can be multiplied only if the number of columns of the first matrix equals the number of rows of the second matrix.
In general, if A is of order m × n and B is of order n × p then A × B exists and its order is m × p. Such a matrix is said to be conformable where m × n multiplied by n × p results in a matrix of order m × p.
The order of AB should be established before multiplying.
The procedure for multiplying two 3 × 3 matrices is outlined below.
If and
then
Position
House 1st 2nd 3rd
Hamilton 60 63 51
Leslie 71 64 74
Barnes 64 69 71
Cunningham 69 72 68
Position Points
1st 2nd 3rd
5 3 1
5 3 1
A
60 63 51 71 64 74 64 69 71 69 72 68
= B
5 3 1
= C
540 621 598 629 =
A
a11 a12 a13 a21 a22 a23 a31 a32 a33
= B
b11 b12 b13 b21 b22 b23 b31 b32 b33 =
The rows of the first matrix are multiplied by the columns of the second matrix. The sum of the products of the elements of row 1 multiplied by column 1 results in the row 1, column 1 element.
The sum of the products of the elements of row 3 multiplied by column 2 results in the row 3, column 2 element.
AB =
a11×b11+a12×b21+a13×b31 a11×b12+a12×b22+a13×b32 a11×b13+a12×b23+a13×b33 a21×b11+a22×b21+a23×b31 a21×b12+a22×b22+a23×b32 a21×b13+a22×b23+a23×b33 a31×b11+a32×b21+a33×b31 a31×b12+a32×b22+a33×b32 a31×b13+a32×b23+a33×b33
,
a Write down the order of the two matrices. b Which of these products exist? i AB ii BA
c Write down the order for the products which exist. d Calculate the products which exist.
THINK WRITE
a Matrix A has 2 rows and 3 columns.
a A is a 2 × 3 matrix.
Matrix B has 3 rows and 2 columns.
B is a 3 × 2 matrix.
b i AB is the product of a 2 × 3 and a 3 × 2 matrix so it exists. A and B are conformable.
b i AB exists since a 2 × 3 matrix multiplied by a 3 × 2 matrix results in a 2 × 2 matrix.
ii BA is the product of a 3 × 2 and a 2 × 3 matrix so it also exists.
ii BA exists since a 3 × 2 matrix multiplied by a 2 × 3 matrix results in a 3 × 3 matrix.
c i The product of AB is a 2 × 2 matrix.
c i AB is a 2 × 2 matrix.
ii The product of BA is a 3 × 3 matrix.
ii BA is a 3 × 3 matrix.
d i Multiply the rows of matrix A by the columns of matrix B.
d i AB =
AB =
Simplify AB. AB=
A 1 2 3
4 5 6
= B
2 –1 0 4 5 3 =
1
2
1 1 2 3
4 5 6 2 –1 0 4 5 3
1×2+ 2×0+ 3×5 1×–1+ 2×4+ 3×3
4×2+ 5×0+ 6×5 4×–1+ 5×4+ 6×3
2 17 16
38 34
5
C h a p t e r 3 M a t r i c e s
135
Note: In worked example 5, AB is a 2 × 2 matrix but BA is a 3 × 3 matrix. In general, matrix multi-plication is not commutative.
That is, for two matrices A and B, AB ≠ BA.
For the product AB we say that A is post-multiplied by B and B is pre-multiplied by A.
The identity matrix
There is one circumstance in which matrix multiplication is commutative. Look at the following example.
This example demonstrates the only case in which matrix multiplication is always com-mutative — that is, when AI = IA = A. Here, I is called the multiplicative identity matrix.
THINK WRITE
ii Multiply the rows of B by the columns of A.
ii BA=
ii BA=
Simplify BA. ii BA=
1
2 –1 0 4 5 3
1 2 3 4 5 6
2×1+–1×4 2×2+ 1– ×5 2×3 + 1– ×6 0×1+4×4 0×2+4×5 0×3+4×6 5×1+3×4 5×2+3×5 5×3+3×6
2
2 – –1 0 16 20 24 17 25 33
If and , calculate AI and IA.
THINK WRITE
A and I are both 2 × 2 matrices so both the products AI and IA exist and are of order 2 × 2.
A 2 × 2 matrix multiplied by a 2 × 2 matrix results in a 2 × 2 matrix.
Find AI using the procedure for multiplying matrices.
Find IA using the procedure for multiplying matrices.
A 2 –3
5
– 4
= I 1 0
0 1 =
1
2 AI 2 –3
5
– 4
1 0 0 1 =
2 –3 5
– 4
=
3 IA 1 0
0 1
2 –3 5
– 4
=
2 –3 5
– 4
=
6
WORKED
E
xample
The multiplicative identity matrix, I, acts in a similar fashion to the number 1 when numbers are multiplied, where I is the multiplicative identity matrix).
An identity matrix can be defined only for square matrices; that is, for matrices of order 1 × 1, 2 × 2, 3 × 3. The other feature of an identity matrix is that it has the number 1 for all elements on the leading diagonal and 0 for all other elements.
AI = IA = A where A is a square matrix and I is the multiplicative identity matrix. If A is not square (say it is 3 × 2), then A × I = A means I would have to be a 2 × 2 matrix because a 3 × 2 matrix multiplied by a 2 × 2 matrix results in a 3 × 2 matrix.
But I × A = A means that I would be a 3 × 3 matrix because a 3 × 3 matrix multiplied by a 3 × 2 matrix results in a 3 × 2 matrix. However, I cannot be a 2 × 2 and a 3 × 3 matrix at the same time. Therefore I can be defined only for square matrices.
Multiplying matrices
1 , , , , ,
a Write down the order of the six matrices. b Which of the following products exist?
c Write down the order of the products which exist. d Calculate those products which exist.
2 a If and , calculate MN and NM.
b Is matrix multiplication commutative? That is, does MN = NM?
3 , , , , and
Calculate the following products.
a AB b AC c DO d DI e IB f BC g CD h CA i OI j ID
i AC ii CA iii DB iv BD v AE vi AI
vii IA viii IB ix EB x E2 xi A2 xii EC
Leading diagonal
0 0
0 0
1 0 0
0 1 0
0 0 0
0 0
0 0
1 0
0 1
0 1 0 0
remember
1. In general, if A is of order m×n and B is of order n×p then A×B exists and its order is m×p; that is, A and B are conformable.
2. In general, for two matrices A and B, AB≠BA.
3. AI=IA=A where A is a square matrix and I is the multiplicative identity matrix.
remember
3B
WORKED
Example
5
A 2 –3 4 5
= B 1 1
1 0
= C
2 4 6 8 0 1
= D = –2 4 E –2 3 –1 0 –4 2
= I 1 0
0 1 =
Mathc ad
Multiplying matrices
M 2 4
1 – 3
= N 5 2
0 4 =
WORKED
Example
6
A 2 –1 0 3
= B 2 0
0 –3
= C 5 –2
8 3
= D 3 2
8 – 5
= I 1 0
0 1
= O 0 0
C h a p t e r 3 M a t r i c e s
137
4 a Calculate the following products.
i ii iii
b What do you notice about all of the answers?
c What term could be given to these matrices?
5
Use the matrices below to answer questions a to d.
A= , B= , C= , D= , E= , F=
a Which one of the following products does not exist?
b The order of the matrix BD is:
c Which one of the following products gives a matrix of order 2 × 2?
d Which one of the following represents the matrix CE:
6 The matrix below shows the number of wins, draws and losses for two soccer teams, the Sharks and the Dolphins.
Thus the Sharks have 10 wins, 2 draws and 5 losses. If 3 points are awarded for a win, 1 for a draw and 0 for a loss:
a write down a 3 × 1 matrix for the points awarded
b use matrix multiplication to find the total points for the two teams.
7 In Australian Rules Football, 6 points are awarded for a goal and 1 point for a behind. The scores in two games were:
Southport 18–12 defeated Broadbeach 14–15 and Lions 10–14 defeated Eagles 9–16.
A AD B AB C BC D FC E CE
A 2 × 2 B 3 × 3 C 2 × 3 D 5 × 3 E 4 × 3
A BF B AB C DC D BC E FD
A B C Does not
exist.
D E
4 3 5 4
4 –3 5
– 4
2 – –3
5 – –8
8
– 3
5 –2
1 – –2
2 – –5
5
– 2
2 –1
m
multiple choiceultiple choice
3 2 0 1
2 –2 4
1 3 6
2 5
1 –3
0 4
1 2 3
2
– 0 2
4 1 –3
5 2
1
– –3 3 –2 4
20 –17 1
– 8
8 –12
5 –11
8 11
4 – –12
5 8 –4
11
– 11 –12
20 –1 8
17
– 8 –12
10 2 5 8 7 2
The first number is for goals scored and the second is for behinds.
a Write the results in a 4 × 2 matrix.
b Write down the 2 × 1 matrix for the points.
c Use matrix multiplication to find the total number of points scored by each team.
8 Two shops, A and B, are supplied with boxes of different brands of chocolates — Yummy, Scrummy and Creamy — as shown in this table:
The cost of the boxes are Yummy $10, Scrummy $25 and Creamy $12.
a Write down the costs in a 3 × 1 matrix.
b Use matrix multiplication to find the total cost for each shop.
Yummy Scrummy Creamy
Shop A 20 20 10
Shop B 10 5 10
Work SHEET
3.1
History
of mathematics
O L G A TA U S S K Y- T O D D
( 3 0 A u g u s t 1 9 0 6 – 7 O c t o b e r 1 9 9 5 )
During her life …
Mt Everest is finally climbed.
The Richter scale for measuring the strength of earthquakes is devised.
Morse code is first used by the Titanic when it sinks.
Gandhi struggles to free India from British Rule.
Olga Taussky-Todd worked in the fields of matrix theory and number theory.
She was born in Olmütz, now part of the Czech Republic, but when she was three the family moved to Vienna and later to Linz. Her father died early so it became difficult for her to continue her studies. Her father, an industrial chemist, had encouraged her studies in mathematics.
Olga went to the University of Vienna where she studied mathematics and
chemistry. She completed a doctorate in 1930 with research into algebraic number fields. After completing her studies she was employed at the university of Göttingen as an assistant and worked with Helmut Ulm by editing his book on number theory. By 1932 Olga had been promoted to the position of tutor.
In 1935 Olga moved to Cambridge where she undertook a research fellowship before
moving to London in 1937 to take up a teaching position. In London she met and later married Jack Todd.
After the Second World War, the couple moved to America where Olga began work on the design of computers. In 1943, she moved to the Ministry of Aircraft where she
conducted research into stability in matrices. This work encouraged her to look in more detail at matrix theory.
Olga was awarded the Austrian Cross of Honour, which is Austria’s highest award; in 1964, she was named woman of the year by
the Los Angeles Times. In 1970 she was
awarded the Ford Prize for her publication on ‘The Sums of Squares’. In 1971 she was named Professor Emeritus at CalTech.
Questions
1. What field of mathematics was Olga’s speciality?
2. What did Olga work on when she moved to America?
3. What award did Olga receive from the Austrian Government?
4. Where was Olga a professor?
Research
C h a p t e r 3 M a t r i c e s
139
Powers of a matrix
A logical extension of matrix multiplication is using the power of a matrix, where A1 = A
A2 = AA
A3 = A2A, and so on.
In general form, An = An –1A, where n is a positive integer. But what dimension can matrix A have?
Matrix powers
Investigate powers of matrices by completing the following steps. (Remember to use pronumerals for the elements of A, not constant values.)
1 a Let matrix A be any 3 × 2 matrix. b Find A2.
c What do you notice?
2 a Let matrix A be any 2 × 2 matrix. b Find A2.
3 a Let matrix A be any 2 × 3 matrix. b Find A2.
c What do you notice?
4 a What general conclusion can you make concerning the order of a matrix that is to be raised to a power?
b Justify your conclusion by referring to the dimensions of matrices involved in a product.
From the above investigations we can conclude that A × A must be conformable; that is, the number of columns of the first factor in the product should be the same as the number of rows in the second factor.
That is, A must be a square matrix where n × n is multiplied by n × n to get A2. Hence powers of matrices are only defined for square matrices.
If A= , find: a A2 b A3
Continued over page
THINK WRITE
a Write the power as a product. a A2=A×A
A2=
A2= 1 3
1
– 2
1 3 1
– 2
1 3 1
– 2
2
– 9
3
– 1
7
WORKED
E
xample
Powers of a matrix
1 If A= , find:
a A2 b A3 c A4
2 If A= , find:
a A2 b A3 c An
3 If A= , find:
a A2 b A3
4 If A= :
a find A2 b confirm that A2A=AA2
THINK WRITE
b Write the power as the product of lesser
powers.
b A3=A2A
A2=
A2= 2
– 9
3
– 1
1 3 1
– 2
11
– 12
4
– –7
remember
1. The power, n, of matrix A, in general form, is An=An – 1A, where n is a positive integer.
2. Powers of matrices are only defined for square matrices; that is, A has to be a square matrix to obtain An.
remember
3C
W WORKEDORKED
E Examplexample
7
2 –1
0 0
1 0 0 0 1 0 0 0 1
1 0 0 2 3 0 0 1 1
1 1 0
2
– – 2 –1
C h a p t e r 3 M a t r i c e s
141
Applications of matrices
1 A garden supplier provides live plants for displays
in 5 penthouse patios, 7 office foyers, 3 banks and 4 hotels. The plants in each different type of display are listed below.
• The patio displays consist of 2 ferns, 1 camellia, 1 geranium and 2 hanging baskets.
• The office foyer displays have 1 palm, 1 geranium, 3 hanging baskets and 2 indoor plants.
• The bank displays have 1 palm, 3 camellias and 5 indoor plants.
• The hotel displays have 2 palms, 3 ferns, 2 camellias, 2 hanging baskets and 5 indoor plants.
The wholesale cost of each plant bought by the supplier is: ferns $22, palms $18, geraniums $8, camellias $15, hanging baskets $12 and a variety of indoor plants that cost $10 on average.
The supplier needs to be able to use this information to calculate costs of displays, number of plants required and profits, but in this form, the information is difficult to handle.
a Develop matrices to display the following information (labels outside the
matrices will help clarify the meaning of the elements):
iii the number of displays supplied to each type of venue
iii the number and variety of plants used in each display
iii the cost of each type of plant.
b Use matrix operations to determine the following:
iii the quantities of each plant needed to fill the orders
iii the supplier’s total outlay to provide the displays
iii the charge for each type of display if the supplier adds 80% profit to the
(iii) cost.
2 A home builder advertised three
designs of ‘Ownit Homes’ to entice people to buy rather than rent their home — the ‘Taps’ for $129 per week, the ‘Avalon’ for $169 per week and the ‘Torana’ for $198 per week. The weekly payments were based on finance
available from a public finance company.
Ownit Homes received orders for 10 ‘Taps’ homes, 8 ‘Avalon’ homes and 12 ‘Torana’ homes. The materials (given in units as stated in their building guide) required for each home are listed below:
• The ‘Taps’ home requires 9 units of steel, 11 of timber, 6 of glass, 7 of paint and 20 of labour.
• The ‘Avalon’ home requires 12 units of steel, 14 of timber, 15 of glass, 12 of paint and 25 of labour.
• The ‘Torana’ home requires 14 units of steel, 12 of timber, 12 of glass, 16 of paint and 24 of labour.
Multiplicative inverse and solving
matrix equations
In question 4 of exercise 3B, you should have found that the product of the matrices was I. This means that one matrix is the multiplicative inverse of the other. We use the symbol A−1 for the multiplicative inverseof A.
If AA-1=A-1A=I, then A−1 is called the multiplicative inverse of A.
In working with numbers, a similar result would be 7 × = 1 or × = 1. Numbers
such as these are called reciprocals or multiplicative inverses of each other.
To reduce costs all materials are purchased from one supplier. The prices per unit are steel $550, timber $950, glass $850, paint $550 and labour is priced at $940 per unit.
Use matrix methods to obtain the following information:
a the amount of money the bank would receive per week from the repayments on these homes
b the total cost of raw materials for all the constructions.
3 A small bakery sells 5 main items: sugar rolls, bread, cakes, pastry and buns. The major ingredients (given in applicable units) required to make one of each item are listed below.
• Sugar rolls (1 dozen) require 1 egg, 4 units of flour, 0.25 of sugar, 0.25 of shortening and 1 of milk.
• Bread (1 loaf) requires 3 units of flour, 0.25 of shortening.
• Cake (1) requires 4 eggs, 3 units of
flour, 2 of sugar, 1 of shortening and 1 of milk.
• Pastry (1) requires 1 egg, 1 unit of flour, 0.33 of shortening.
• Buns ( 1 dozen) require 2 units of flour, 3 of sugar, 1 of shortening and 1 of milk. Two suppliers (Supplier 1 and Supplier 2) provide quotes for the ingredients, given as ordered pairs:
eggs (10, 12), flour (8, 10), sugar (10, 12), shortening (12, 15) and milk (12, 12). For one office function the following orders were received:
15 dozen sugar rolls, 150 loaves of bread, 45 cakes, 65 pastries and 35 dozen buns.
a Represent all the above information in matrix form taking into account ingredients, orders, suppliers’ quotes.
b Use these matrices to provide a list of the amounts of the ingredients required to fill the orders for the function.
c Which supplier provides the cheapest total quote? What savings are made by using this supplier?
d Provide a list of selling prices if a 90% markup on the cost prices is used to fix the price.
e Calculate the total takings based on this information from part (d).
1 7
--- 4 5 --- 5
---C h a p t e r 3 M a t r i c e s
143
Inverse of a matrix
Consider matrix A, a 2 × 2 matrix, such that A= . If a multiplicative inverse of A exists, then A×A–1=I.
If A–1 exists, let A–1= That is,
AA–1= I LHS =
LHS =
LHS = RHS LHS =
If and , find AB and hence write down the multiplicative inverse
of A.
THINK WRITE
AB will be a 2 × 2 matrix since A and B are both 2 × 2 matrices.
since 6 is a common factor
of each element.
AB= 6I
To produce I we need to multiply both sides by .
Since , the inverse of A is B.
So A−1= B
=
=
A 4 1
6 3
= B 3 –1
6
– 4
=
1 AB 4 1
6 3
3 –1 6
– 4
=
6 0 0 6 =
2 6 0 0 6
6 1 0 0 1
6I
= =
3 16--- A( )16---B = I
4 A( )16---B = I 1
6
---1 6
---3 6 --- 1
6
---–
6 6
---– 4
6
---1 2 --- 1
6
---–
1
– 2
3
---8
WORKED
E
xample
a b c d x y
u v
a b c d
x y u v ax+bu ay+bv cx+du cy+dv
1 0 0 1
Equating terms of the two matrices: ax+bu= 1 [1] cx+du= 0 [2] ay+bv= 0 [3] cy+dv= 1 [4]
Solving this system of simultaneous equations:
Use [1] and [2] to eliminate x by multiplying equation [1] by c and equation [2] by a. acx+bcu= c [5]
acx+adu= 0 [6]
Equation [6] minus equation [5] gives: adu−bcu = −c
u(ad−bc) = −c
u = This will replace u in A–1. Continue in a similar fashion to arrive at:
y = x = and v =
Therefore A–1 =
= where ad−bc≠ 0
If ad – bc= 0 then this scalar is undefined, therefore A–1 does not exist. That is, there is no matrix that, when multiplied by A will yield I, the identity matrix. If A has no inverse then it is said to be singular.
There is a relationship between A and A−1 which is outlined below. If A is the matrix , proceed as follows.
1. Swap the elements on the main diagonal of A and multiply the elements
on the other diagonal by −1 . This gives the matrix .
2. Find the product of A and this matrix:
=
3. This will equal I if we divide each element by (ad − bc) (or multiply by ). These steps demonstrate a clear method for finding the multiplicative inverse of a matrix.
c – ad–bc
---b – ad–bc
--- d ad–bc
--- a ad–bc
---d ad–bc --- –b
ad–bc
---c – ad–bc --- a
ad–bc
---1 ad–bc
--- d – b c
– a
a b c d
d –b c
– a
a b c d
d –b c
– a
ad–bc 0 0 ad–bc
1 ad–bc
( )
---C h a p t e r 3 M a t r i c e s
145
The inverse of is .
The number (ad−bc) is called the determinant of the matrix A and is written as det A or |A|.
Note:Only square matrices have inverses.
We will be concerned only with the inverse of 2 × 2 matrices at this stage in this course.
We can check that and .
C–1 is the multiplicative inverse of C if C×C–1=I.
Singular matrices
Matrices for which the determinant equals 0 do not have an inverse, since is undefined. Such matrices are called singular matrices.
If det A= 0 then A is singular and an inverse does not exist.
A a b
c d
= A–1 1
ad–bc --- d –b
c
– a
=
If find C−1.
THINK WRITE
Write the general form of C and the general form of its inverse.
C= C–1=
Swap the elements on the main diagonal of C.
Multiply the elements on the other diagonal of C by −1.
C–1=
=
Write down the inverse of C. C–1 =
C 2 –3
1 5 =
1 a b
c d
1
ad–bc
--- d –b
c – a 2 5 2 3 1 – 1 2×5
( )–(–3×1) --- 5 3
1
– 2
1 10–( )–3 --- 5 3
1
– 2
3 13---1 5 3
1
– 2
9
WORKED
E
xample
CC–1 = I C–1C = I
CC–1 1 13 --- 2 –3
1 5 5 3 1 – 2 = 1 13 --- 13 0
0 13 =
I =
C–1C 1 13 --- 5 3
1 – 2
2 –3 1 5 =
1 13 --- 13 0
0 13 = I = 1 0
Further matrix equations
Matrix equations of the type AX = B may be solved by using the properties of multiplicative inverses.
A matrix equation AX = B is similar to the equation 3x = 7. To solve this we would divide both sides of the equation by 3 (or multiply by ). To solve the matrix equation we multiply both sides by A−1. Since the order of multiplying matrices is important we must be careful in which position we multiply by the inverse.
1. For AX = B
Pre-multiply by A−1: A−1AX = A−1B
or IX = A−1B since A−1A = I. X = A−1B since IX = X. 2. For XA = B
Post-multiply by A−1: XAA−1= BA−1
or XI = BA−1 since AA−1= I X = BA−1 since XI = X.
1. If AX = B, then X = A−1B. 2. If XA = B, then X = BA−1.
Note: A–1 cannot be ‘inserted’ between 2 matrices. It can either pre- or post-multiply A on one side of a matrix equation.
1 3
and
Find X if:
a AX=B b XA=B.
THINK WRITE
a We require A−1 so first calculate det A. a A= ∴ det A= 3 − 0 = 3
Write down A−1. A−1=
Write the equation. AX=B
Pre-multiply both sides of the equation
by A−1. A−1AX=A−1B
Remember A−1A = I and IX = X.
Calculate the product of A−1 and B. X=
=
A 1 2
0 3
= B 2 5
2 – 1 =
1 1 2
0 3
2 13--- 3 –2
0 1
3 4
5
6 13--- 3 –2
0 1 2 5 2 – 1
1 3 --- 10 13
2
– 1
10
C h a p t e r 3 M a t r i c e s
147
In part a of worked example 10 both sides of the equation were pre-multiplied by A−1; in part b both sides were post-multiplied by A−1. Remember that the matrix and its inverse must be next to each other so that AA−1= I.
Fractional scalars should be left outside the matrix unless they give whole numbers when multiplied by each element.
Multiplicative inverse and
solving matrix equations
1 If and , find AB and hence write down:
a the inverse of A b the inverse of B.
2 If and , find MN. Hence write down M−1 and N−1.
THINK WRITE
b Write the equation. b XA=B
Post-multiply both sides of the equation by A−1.
X=BA−1
Calculate the product of B and A−1
using A−1 which was found in part a. =
=
1
2
3 13--- 2 5
2 – 1
3 –2 0 1
1 3 --- 6 1
6 – 5
remember
1. If AA−1=A−1A=I, then A−1 is called the multiplicative inverse of A.
2. The inverse of A= is A−1=
The number (ad−bc) is called the determinant of the matrix A and is written as det A or |A|.
3. If det A= 0 then A is singular and an inverse does not exist. 4. (a) If AX=B, then X=A−1B.
(b) If XA=B, then X=BA−1. a b c d
1
ad–bc
--- d –b c
– a
remember
3D
SkillS HEET
3.2
Mathca
d
Inverse matrices WORKED
Example
8
A 4 1
2 – 1
= B 1 –1
2 4 =
M 2 6
0 –1
= N –1 –6
0 2 =
3 Calculate the determinants of the following matrices.
4 Write down the inverses of each matrix in question 3.
5
Using the matrices below, select the correct answer in questions a to d.
, and
a Det P is equal to:
b R−1 is equal to:
c Det (PQ) is equal to:
d If QX = R, then X is equal to:
6 Write down a 2 × 2 matrix which is singular.
7 ,
Find:
8 Explain why these matrices do not have an inverse.
a b c
d e f
A 10 B 2 C −10 D −2 E 8
A − B C
D − E
A −30 B 10 C −10 D 30 E −20
A − B C −
D E −
a C−1 b D−1 c CD d (CD)−1 e C−1D−1 f D−1C−1
a b c
A 2 3
5 10
= B –2 –3
4 0
= C 2 6
0 –1 =
D 4 3
4 1
= E –2 –1
3 – –5
= F 2 –1
6 4 = WORKED Example 9 m
multiple choiceultiple choice
P 4 3
2 –1
= Q 2 –3
1
– 0
= R 8 6
4
– 2
=
1 8 --- 2 –6
4 8
1 8
--- 8 6
4
– 2
1 40 --- 8 6
4 – 2
1 40 --- 2 –6
4 8
1 40 --- 2 –6
4 8
1 3
--- –12 6
0 10
1 3
--- 12 6
0 –10
1 3
--- 12 6
0 –10
1 3 --- 6 28
2 8
1 3 --- 6 28
2 8
C 2 6
0 –1
= D 0 –2
2 1 =
D 2 1
4 2
= E –2 –4
C h a p t e r 3 M a t r i c e s
149
9 If and , find:
10 Let , , . Find X if:
11 , and X = .
Solve these matrix equations.
12 Find the value of x and y by solving these matrix equations.
The transpose of a matrix
The transpose of matrix A is A′, where A= and A′= . The transpose of
a matrix is an interchange of rows and columns (row 1 becomes column 1 and so on).
Consider the following laws that apply to the transpose of matrices A and B: 1. (A¢)¢=A
2. (A + B)¢=A¢+B¢
3. (kA)¢=kA¢
4. (AB)¢=B¢A¢
5. AA¢ is a symmetric matrix (i.e. A¢=A).
The proofs of these laws are given as problems in exercise 3E.
a AB b (AB)−1
a AX=B b XA=B c XC=A d AX=C
e ABX=C f CX=C g XB=I h A−1BX=C
a b
a b
c d
A 4 0
1
– 1
= B 0 2
1 – 0 = W WORKEDORKED E Examplexample
10
A 2 3
4 5
= B 5 5
2
– 1
= C 0 –1
6 6
=
A 3 4
1
– 2
= B 6 1
2 1 = x y AX 2 4 –
= BX 15
7 =
Work SHEET
3.1
3 4 1 – 5 x y 2 – 7
= 2 3
4 –1 x y 8 2 = 4 – 2
3 –2 x
y
14 12 –
= –1 3
2 –3 x y 5 2 = a b c d a c b d
remember
1. When required to prove a statement is true:
(a) do not assume it is true and use the statement in your proof (b) work only one side of the statement at a time, not both together
(c) do not use actual constant values for the elements, use pronumerals only. 2. If you are asked to show a statement is true, you are expected to use actual
values as given.
remember
The transpose of a matrix
1 Prove that for any 2 × 2 matrix A, (A′)′=A.
2 Show that for A = and B = , (A+B)′=A′+B′.
3 Show that for k=−2 and A= , (kA)′=kA′.
4 Show that for A= and B= , (AB)′=B′A′.
5 Show that for any 2 × 2 matrix, AA′ is symmetrical.
Applications of matrices
Application 1: Simultaneous equations
As we saw in questions 11 and 12 from exercise 3D, matrices may be used to solve linear simultaneous equations. The pair of equations may be written in the form AX=B
where A is the matrix of the coefficients of x and y in the equations, X= and B is
the matrix of the numbers on the right-hand side of the simultaneous equations.
A is called the coefficient matrix.
For example, the simultaneous equations:
ax+by=e
cx+dy=f
can be expressed as the matrix equation:
which is of the form AX = B. Here is called the coefficient matrix, the
variable matrix and the constant matrix.
As we have seen, this equation can be solved by using:
A−1AX=A−1B X=A−1B
3E
1 3
0 2
0 –1
1 2
3
– 1
0 1
3 –4
1 2
0 1
1 0
x y
a b
c d
x y
e f
=
a b
c d
x y
C h a p t e r 3 M a t r i c e s
151
Most graphics calculators provide a facility for calculating inverses of matrices. To solve the equations in worked example 11 using the TI-83 graphics calculator, follow these steps.
1. Press .
2. Choose the EDIT menu, select [A] and press .
3. Specify that [A] is a 2 × 2 matrix. Solve 3x−y= 16 and 2x+ 5y= 5 by matrix methods.
THINK WRITE
Write the simultaneous equations under each other making sure the variables are in corresponding positions.
3x−y= 16 2x+ 5y= 5
Write the matrix equation. AX=B
=
Rearrange the equation in general form so that X is the subject.
A–1AX=A–1B IX=A–1B X=A–1B
Calculate A−1. A–1=
Multiply A–1 by B. X=
=
Write the answers in the form x = . . . and y = . . .
Note: The solution should be verified by substituting x = 5 and y = −1 into the original equations.
∴x= 5 and y= −1
1
2
3 –1
2 5
x y
16 5
3
4 1
17
--- 5 1 2
– 3
5 1
17
--- 5 1 2
– 3
16 5
1 17 --- 85
17 –
x y
5 1 – =
6
11
WORKED
E
xample
Graphics Calculator
Graphics Calculator
tip!
tip!
Solving matrix equations
MATRX
ENTER
4. Let A be the first matrix in step 1 of worked example 11, and let B be the last matrix. Insert each value into [A] by typing the number and then pressing .
5. Go back to and the EDIT menu and enter the values for [B].
6. Use to return to the home
screen.
7. Press .
8. Choose the NAMES menu and select [A].
9. Press to produce [A]−1.
10. Press , select [B] from the NAMES menu.
11. Press to calculate [A]−1[B]. You should find that x = 5 and y = −1 as before.
In some cases a matrix may have elements with a large number of decimal places. In such cases you can scroll across a row using the scroll key .
The MATH menu provides a number of matrix operations.
Use 1:det to find the determinant of a matrix
by pressing MATH 1 NAMES
[choose the matrix].
Use 5:identity to find the identity matrix by pressing MATH 5 and entering a number. (For example, 3 will give the 3 × 3 identity matrix.)
Application 2: Summarising information
We have already seen how matrices may be used to summarise information such as town–road connections. Information which can be summarised in tabular form may also be presented as a matrix.
ENTER
MATRX
2nd [QUIT]
MATRX
x−1
MATRX
ENTER
▼
MATRX MATRX
MATRX
In a large country town, there are three major supermarkets. Customers switch from one to another due to advertising, better service, prices and for other reasons. A survey of 1000 customers has revealed the following information for the past month.
Best Buys started with 40% of the market; 90% of its customers remained loyal to Best Buys but 5% changed to Great Groceries and 5% to Super Store.
Great Groceries started with a 36% market share; 85% remained loyal, 10% transferred to Best Buys and 5% to Super Store.
Super Store started with 24% of the customers; it lost 15% to Best Buys and 5% to great Groceries, but 80% remained.
Summarise the information in matrix form and calculate the new market shares.
12
C h a p t e r 3 M a t r i c e s
153
THINK WRITE
The information may be
summarised in a 3 × 3 matrix with the rows representing retention rates and gains and the columns representing retention rates and losses. This may be called a transition matrix.
Row 1 shows that Best Buys retains 90% of its customers, gains 10% of Great Groceries’ customers and gains 15% of Super Store’s customers. Column 1 indicates that Best Buys retains 90% of its customers, loses 5% to Great Groceries and loses 5% to Super Store. Note that each column totals 100%.
Write the initial market shares as a 3 × 1 matrix. This information is found as the market share at the beginning of the month. Note: The values total 1.
The initial market share matrix is
The new market share will be the transition matrix, converted to decimal numbers, multiplied by the market share matrix.
Express the new market shares as percentages. Check the values add up to 100%.
The new market shares are Best Buys 43.2%, Great Groceries 33.8% and Super Store 23.0%.
1
Retention rates and losses (%)
Best Buys
Great Groceries
Super Store
Best Buys
90 10 15
Great Groceries
5 85 5
Super Store
5 5 80
Retention rates and gains
(%)
2
0.40 0.36 0.24
3
0.90 0.10 0.15 0.05 0.85 0.05 0.05 0.05 0.80
0.40 0.36 0.24
0.432 0.338 0.230 =
4
remember
1. Matrices may be used to solve simultaneous equations:
ax+by=e
cx+dy=f.
The pair of equations may be written in the form AX=B, where A= ,
X= and B= .
2. Matrices can also be used to summarise information which is in table form and solve related problems; however, care must be taken in setting up the matrices.
a b
c d
x y
e f
remember
Applications of matrices
In the following exercise solve all problems manually then use a graphics calculator wherever appropriate to check your solutions.
1 Solve these simultaneous equations by matrix methods.
2 Consider these two pairs of simultaneous equations:
a Show by algebraic means that the simultaneous equations in i have no solution.
b Show that the simultaneous equations in ii have an infinite number of solutions.
c Write the equations in matrix form and explain how these facts are related to the
determinant of the matrix of the coefficients.
d Draw, on two sets of axes, graphs of the two lines in each of i and ii.
e Explain how the graphs are related to parts a and b.
3
Consider the simultaneous equations: 3x− 2y= 5
y+ 2x= 8
a The coefficient matrix is:
b The solution to the simultaneous equations is:
4
In an alternative Australian Rules Football game, a team gains x points for a goal and y
points for a behind. In one game Cairns obtained 66 points by scoring 10 goals and 8 behinds and Townsville obtained 70 points from 12 goals and 5 behinds.
a This information is represented by which of the following matrix equations?
b The value of x−y is:
a 2x− 3y= 13 and x+ 2y= 3 b 3x+y= 9 and −2x+ 5y= −6
c −x+ 4y= −2 and x− 5y= 0 d 6x+ 7y= 0 and 4x− 3y= 0
e 4x+y= 20 and x−y= 0 f 3x− 2y= 0 and x−y= 1
i 3x− 2y= 4
6x− 4y= 12
ii 3x− 2y= 6
6x− 4y= 12
A B C D E
A x= 2, y= 3 B x= 3, y= 2 C x= , y=
D x= −2, y= 2 E x= , y=
A B C
D E
A 5 B 4 C 6 D 3 E 2
3F
SkillS
HEET
3.3
WORKED Example
11
Mathc ad
Matrices and simultaneous equations
m
multiple choiceultiple choice
3 –2
1 2 5 8 3 2 2 – 1 3 1 2 – 2
3 –2
2 1 13 4 --- 19 8 ---4 3 --- 7 3 ---m
multiple choiceultiple choice
8 10 5 12 x y 66 70
= 10 12
8 5
x y
66 70
= 10 8
12 5 x y 66 70 = 8 5 10 12 x y 66 70
= 12 10
C h a p t e r 3 M a t r i c e s
155
5 The sum of two numbers is 20 and their difference is 12. Find the numbers by setting
up simultaneous equations and solving by matrix methods.
6 In a factory, two types of components are processed on two separate machines. The
respective processing times on the first machine are 18 minutes and 21 minutes, while for the second machine the times are 4 minutes and 42 minutes. How many of each type of component, per machine, should be processed in an 8-hour shift so that both machines are fully occupied and the output of each machine is the same?
7 In a swimming competition, 5 points are awarded for first place, 3 for second, 2 for
third and 1 point for an unplaced result. The top competitors’ results were:
Place the results and points in suitable matrices and use matrix multiplication to find the highest points scorer.
8 Cyril’s circus arrived in town last week and during the
week the number of adults, children and pensioners attending the circus was recorded for the first five shows (see table below).
The entry cost is $20 for adults, $6 for children and $5 for pensioners.
Set up the information in suitable matrices to find the total takings for the first five shows.
Name
No. of races competed in
First placings
Second placings
Third placings
Rania 6 2 — 2
Patricia 4 4 — —
Anh 5 3 2 —
Mayssa 6 1 3 2
Rachel 6 2 3 —
Adults Children Pensioners
Monday 400 200 20
Tuesday 450 350 50
Wednesday 370 410 45
Thursday 290 380 70
Friday 420 530 65
WORKED
Example
12
Work
SHEET
3.2
Matrix multiplication
using a graphics calculator
Worked example 12 may be solved using a graphics calculator as follows. Enter the 3 × 3 transition matrix as matrix A:1. Press , select EDIT and 1:[A]. 2. Change the dimensions of A to 3 × 3 and
press .
3. Enter the values from worked example 12 as shown.
Enter the 3 × 1 market share matrix as matrix B: 4. Press , select EDIT and 2:[B]. 5. Change the dimensions of B to 3 × 1 and
press .
6. Enter the values from worked example 12 as shown.
Now multiply matrices A and B:
7. Press [QUIT] to return to the home screen. 8. Press , select NAMES and 1:[A], and
press . 9. Press the ¥ key.
10. Press , select NAMES and 2:[B], and press .
11. Press to perform the matrix multiplication.
Questions
Use a graphics calculator to find A × B for each of the following:
1 A = , B =
2 A = , B =
3 A = , B =
MATRX ENTER
MATRX ENTER
2nd MATRX ENTER
MATRX ENTER ENTER
1 3 3 5 0 2 6 2 8
9 2 5
2.5 6.1 9.2 0.3 6.6 0.7 3.7 4.6
3.7 0.4 9.4 5.1 5.9 2.2
2 3 0 –2 7 0 1 –1 4 –7 5 –3 9 –2 6 5
– 1 1 9 6
4 4 4 4 4
3
– 8
5 11 4 –7 1 – –2
C h a p t e r 3 M a t r i c e s
157
Dominance matrices
Have you ever wondered how tennis players are seeded or ranked? It obviously has something to do with their performance against past opponents. In a knock-out compe-tition, one loss and you are out of the competition. Only the winners continue to play. Dominance matrices are often used to determine player rankings.
The investigation below will explain how matrices are used to establish the seedings or rankings of players in round-robin situations where each player plays every other player, thereby creating a more just system of ranking.
Dominance matrices — another
application of matrices
Consider 4 players Alan, Brian, Carlo and Denis (A, B, C, D), who on past performances have shown that A defeats D and B, D defeats B, C defeats A and D, and B defeats C.
This situation can be represented on a digraph — a network diagram that has
arrows on the edges, where A → B indicates that A defeats B.
A
B C
The information