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107Appendix D Matrices and 3 3 Determinants
D1 Matrices: introductionMatrices can be used, among other purposes, to solve simultaneous equations provided we
define their operations (e.g. addition and multiplication) in special ways. The interest in and use
of matrices has increased greatly since the introduction of computers because their operations
are easy to program on a computer. Manually, for example, it is usually quicker to solve
simultaneous equations using determinants. However, with a computer, matrices are much
easier to use, regardless of how many variables are involved.
Definitions
A matrix is a set of numbers (called elements) arranged in a rectangular pattern (or array) of
rows and columns. A determinant , as we have seen in Appendix C, is such an array
distinguished by vertical bars at each side. To distinguish a matrix , the array is enclosed in
parentheses, either round or square.
For example, is a determinant , which has the value 5, but or is a matrix ,
which does not have a value. A matrix does not represent a number.
If a matrix has a rows and b columns, it is said to be an ‘a b matrix’ or to ‘have an order of
a b’. A matrix of order ‘a 1’ is called a column matrix . A matrix of order ‘1 b’ is called
a row matrix .
Examples
1 A is a matrix of order 3 2.
2 D is a matrix of order 4 1, a column matrix .
5
3
1
2
3 0
2 4
5 3
2 1
3 4
2 1
3 4
2 1
3 4
Matrices and 3 3
determinants
D
A PPENDIX
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3 K (5 –3 2) is a matrix of order 1 3, a row matrix.
4 P is a 2 2 matrix, a square matrix of order 2.
5 T
is a 3 3 matrix, a square matrix of order 3.
We identify a matrix by a capital letter and its order can be shown under this letter. For
example, K32
is a matrix that we call K and its order is 3 2 (i.e. 3 rows and 2 columns).
Exercises D1
Below is a set of matrices that are also referred to in following exercises.
A B C D E F (3 1 2)
G H K 1 Using the above set of matrices, state the order of: a D, b G, c C, d A, e F.
2 Which of the matrices in the above set is: a a 2 3 matrix, b a 3 2 matrix, c a square
matrix, d a row matrix, e a column matrix?
D2 Some definitions and lawsEqual matrices
Two matrices are said to be equal if, and only if, they are identical in every respect—that is, theelements of each are the same and in the same positions.
Example
If , then a 2, b 4, c 1, d 3.2 4
1 3
a b
c d
3 1 2
0 2 3
1 0 2
1 3
2 1
2 1
1 3
3 2
1 3 4
2 5 2
4 1 3
2 1 3
1 4 2
3
1
2
2 13 2
1 0 23 2 1
1 0 3
2 3 1
4 3 0
4 –2
0 3
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The sum or difference of two matrices
These are found by adding or subtracting the corresponding elements of each matrix.
Example
The zero matrix
Example
The zero matrix of is . The zero matrix of is .
Multiplication by a constant
By definition, a matrix is multiplied by a constant by multiplying every element of the matrix
by that constant.
Note : The matrix O is not the number zero but is the matrix of the same order as A, which has
the number 0 for each of its elements.
0 0
0 0
2 –3
0 5
0 0 0
0 0 0
a b c
d e f
For matrix A the zero matrix, O, is defined to be the matrix such that A O O A A (the
law of addition of zero). The letter O is used to denote a zero matrix.
From the definition, it can be seen that A B B A (the commutative law for addition).
Note : Two matrices can be added or subtracted only when they have the same order (i.e. they
must have the same number of rows and the same number of columns; they must have the
‘same shape’). The resulting sum or difference will also have the same order.
2 0 1
3 3 2
3 3 0
5 1 2
5 3 1
2 4 0
c i
f
l
b h
e
k
a g
d
j
g h i
j k l
a b c
d e f
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Example
3
Exercises D2
1 Solve the following matrix equations:
a b c d
2 Write the single matrix equation
as three separate simultaneous equations.
3 Using the set of matrices (A, B, C . . . K) in Exercises D1:
a state which pairs of those matrices can be added or subtracted
b write down the matrix K E
c state the zero matrix of D
d state the zero matrix of E
e write down the matrix 3 A
f write down the matrix 2E 3K
D3 Multiplication of matricesSince a matrix is simply an array (arrangement) of numbers in a rectangular pattern and does
not have a value, we can define the product of two matrices in any way we choose.
For the purposes of explanation, the rows and columns of a matrix will be designated as shown
in the matrix below: the rows being called R1, R2, R3 . . . and the columns C1, C2 . . . .
Any particular element of a matrix can be identified by stating its row and its column.
For example, in the matrix , R1C3 7, R1C1 3 and R2C2 5.
By definition, when two matrices are multiplied, the product is another matrix and regardless of
how many rows and columns the matrices may possess, when the elements in Rn
of the first
C1 C2 C3
↓ ↓ ↓3 6 7
2 5 4
R1 →R2 →
87
9
3 x 2 y – 5 z4 x – 3 y 2 z
5 x 5 y – 3 z
3
5
x y
x – y
3 – x
y 9
x – 1
2 y 3
5
6
x 2
y – 3
7
3
x
y
6 3 0
0 9 6
2 1 0
0 3 2
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matrix are multiplied in succession by the elements of Cm
of the second matrix and these
products are added, this gives element RnC
mof the product matrix.
When put into words this definition seems very complicated but some illustrations and some
practice should enable you to gain facility with this process.
Ignoring all the other rows and columns that may be present:
1 R3 → R3 → In the product matrix, element R3C2 (5 4) (2 0) (7 1) 27
2
You are advised to practise this process until it becomes quite familiar to you. Below are some
exercises to enable you to practise the multiplication of two matrices. You will quickly discover
the benefit of using fingers or a pen to obscure the rows and columns not being used to obtain aparticular element of the product matrix.
Example
When the elements are small you should be able to obtain the product matrix without needing
to write down the intermediate steps.
e.g. 7 3
8 12
2 0
1 3
3 1
2 4
11 13
21 23
4 9
8 15
8 3
16 5
(2 4) (3 1) (2 2) (3 3)
(4 4) (5 1) (4 2) (5 3)
4 2
1 3
2 3
4 5
17 23 29
39 53 67
9 20
27 40
7 16
21 32
5 12
15 24
(1 5) (2 6) (1 7) (2 8) (1 9) (2 10)
(3 5) (4 6) (3 7) (4 8) (3 9) (4 10)
5 7 9
6 8 10
1 2
3 4
27
4
0
1
5 2 7
111Appendix D Matrices and 3 3 Determinants
C2
↓C2
↓
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Exercises D3
1 a You are given the following matrices:
A , B , C , D , E Write down the following matrices:
i A B ii A C iii A D iv A Ev B C vi B D vii B E viii C D
ix C E x D E
b You are given the following matrices:
P , Q , R , S , T Write down the following matrices:
i P Q ii P R iii P S iv P T
v Q R vi Q S vii Q T viii R Six R T x S T
2 As always in algebra, if there is no operation sign between two terms, multiplication is to be
assumed. AK means A K, i.e. matrix A matrix K.
Find the product of the matrices in each case below:
a b c d e f
g
h
i (0 3 2) j (2 3 1)
k l (7 2 5) 6
2
8
5
6
2 1
4 3
2
1
3
1
1
4
1 5
4 26 0
3 0 1
0 2 1
1 2 4
3 0 1
1 2
0 3
2 1 1
4 3 0
0 1 5
0 1 1
1 3 3
5 4 8
6 7 9
1 0
3 2
3 2
1 5
1 0
1 2
0 1 2
1 1 0
2 3 1
2 1 0
1 0 3
3 1 1
2 1 0
2 3 0
1 0 1
1 2 1
0 0 3
2 0 1
1 2 3
2 0 2
0 3 1
1 3 0
0 2 1
2 0 3
–4 2
–6 3
–2 0
–3 0
4 –3
2 –1
3 –2
6 –4
–1 3
–2 4
2 0
0 3
5 3
2 1
1 2
4 3
5 1
3 2
3 1
2 4
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m n (4 5 1) o p
3 If A and B , find a A
2, b B
3.
D4 CompatibilityBy now it has probably become clear to you that multiplication of two matrices is only possible
when there are the same number of elements in any row of the first matrix as there are in anycolumn of the second matrix, i.e. the product A
mn B
pq C exists only when n p, that is,
only when the number of columns in the first matrix equals the number of rows in the second matrix.
When, and only when, this is so, the matrices are said to be ‘compatible’ for this multiplication. If
n p, the matrices cannot be multiplied and are said to be ‘incompatible’ for this operation.
Examples
1 The product M N does exist
and has order 2
5.
2 The product P Q does not exist.
3 The product R T does exist
and has order 4 3.
It is quite common for a product A B to exist but for the product B A not to exist. For
example:
V W exists (and has order 2 3), but
2 3 3 3
W V does not exist.
3 3 2 3
–1 2
2 0
3 1 2
0 2 1
1 3 0
4 0
3 2
0 1
5 0 1
1 4 2
3 2 0
1 3
2 1
2 1
1 3
0
2
3
3
0
2 0
1 3
2 4
113Appendix D Matrices and 3 3 Determinants
compatible
order of product
2 5
M N
3 3
incompatible
2 2
P Q
2 3
compatible
4 3
R T
3 3
order of product
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It is easy to show that A B and B A both exist only for Amn
Bnm
:
A B B A
e.g. and both exist.
2 3 3 2 3 2 2 3
Example
O
You can verify this after studying the next section.
Exercises D4
1 Using the set of matrices in Exercises D1, state whether the given product exists in eachcase (answering ‘yes’ or ‘no’) and, if it does exist, state its order.
a AB b BA c AC d DA e FE f EF
g CF h AE i BD j CD k DC l CE
D5 The identity matrix, I
Exercises D51 Write down the product matrix:
a b 2 Write down the product matrix:
a b a b c
d e f
g h i
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
a b c
d e f
g h i
a b
c d
1 0
0 1
1 0
0 1
a b
c d
0 0
0 0
4 2
6 3
3 2
6 4
Summary
Facts about the product of two matrices:
◗ A B may not exist .
◗ Even if A B does exist, it is possible that B A does not exist.
◗ Even when both products exist, in general, A B B A.
◗ It is possible that, A B O even though neither A nor B is the zero matrix O.
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The principal diagonal of a square matrix is the diagonal that runs from the top left-hand corner to
the bottom right-hand corner. The square matrix, which has the number 1 for each element on the
principal diagonal and all other elements zero, plays a very special role in the theory of matrices.
is called the identity matrix of order 2, and is specified as I2.
is called the identity matrix of order 3, and is specified as I3.
For any square matrix An
(i.e. of order n n), An I
n In A
n.
In
plays the same role in matrix theory as unity does in arithmetic (e.g. 7 1 1 7 7),
and so it is called the unit matrix, Unor the identity matrix, In. We use the latter name and
symbol in this book.
We use capital letters to identify matrices but the capital letter I is reserved for the identity
matrix and the capital letter O is reserved for the zero matrix.
Therefore, non-square matrices do not have an identity matrix.
Exercises D5 (continued)
3 For the following, write down the product matrix if it exists. If it does not exist, write
‘incompatible’.
a b
c d
4 a If A , write down the matrix A.
b If B , write down the matrix B.
c Does have an identity matrix? State the reason for your answer.17 –13 19
34 28 15
17 –13 19
34 28 15
17 –13 19
34 28 15
Remember : an identity matrix is always square .
17 –13 19
34 28 –15
17 –13 19
34 28 15
1 0
0 1
1 2 3
4 5 6
1 2 3
4 5 6
1 0
0 1
2 1 3
1 3 2
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1 2 3
4 5 6
Note : For a matrix An m
which is not square, An m
Im An m
(but Im An m
does not exist) and
In An m
An m
(but An m
In does not exist).
Remember : I stands for the Identity matrix, not the numeral 1.
1 0 0
0 1 0
0 0 1
1 0
0 1
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5 For each of the following matrices, state whether an identity matrix exists (answering ‘yes’
or ‘no’) and, if it does exist, write it down.
a b c d
6 If M , show that a M
2 –I, b M
3 –M.
D6 The inverse matrix, A–1
The numbers 13 and
1
1
3 are said to be multiplicative inverses of one another because in
multiplication one undoes what the other does.
For example, 957 13
1
1
3 957, 7
1
1
3 13 7 .
This occurs because 13
1
1
3 1 and
1
1
3 13 1.
Now we will see how this applies to matrices.
Exercises D6
1 If A and B , write down a the matrix AB, b the matrix BA.
Since both products in the exercise above are the identity matrix, you can probably guess that A
and B are said to be inverses of each other.
The inverse of matrix M is written as M–1
, so you have proved for the above matrices A and B
that AB BA I, that is, that B A–1
and A B–1
.
Exercises D6 (continued)
2 Given that C and D :
a write down the matrix CD
b write down the matrix DC
c what have we proved about the matrices C and D?
3 1 1
1 0 1
1 1 0
1 1 1
1 1 2
1 2 1
Definition : Two matrices, A and B, are said to be inverses of one another (i.e. A B–1
and
B A–1
) if AB BA I.
1 3
2 5
5 3
2 1
0 1
1 0
3 2 5
4 1 6
7 9 8
3 4 26 1 5
2 8
7 3
1 0
3 2
7 5
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3 Given that matrix A pq
has an inverse Br s
:
a what is the order of matrix AB?
b what is the order of matrix BA?
c Since these matrices are inverses of each other, by definition AB BA Inn
Therefore, the orders of AB and BA are both n n. What can you deduce about the
values of p, q, r , s and n?
d Hence, if matrices A and B are inverses of each other, what can you deduce about the
shapes of the matrices A and B?
Most square matrices have an inverse, but not all of them. (Actually it can be shown that all
square matrices have an inverse except those for which the determinant A 0.)
A matrix that has an inverse is said to be invertible.
Exercises D6 (continued)
4 If A B C D E :
a write down the products:i AB ii AC iii AD iv AE
v BC vi BD vii BE
b hence, write down: (i) matrix A–1
(ii) matrix B–1
5 Given that J and K :
a write down the matrix JK
b write down the matrix J–1
3 2 1
0 1 2
0 0 3
1 2 1
0 3 2
0 0 1
3 4 2
2 1 0
1 1 1
1 2 0
0 1 2
1 0 2
1 1 0
1 0 1
0 2 1
1 2 2
2 5 4
3 7 5
2 1 1
1 1 1
2 2 1
Note : You are not required to be able to find the inverse A–1
of a given matrix A, but you should
be able to determine whether or not two given matrices A and B are inverses of each other by
testing whether AB BA I.
Summary
◗ If A B B A I, then A and B are inverses of each other (i.e. A B–1
and B A–1
).
◗ A non-square matrix cannot have an inverse.
◗ Most, but not all, square matrices have an inverse (i.e. they are invertible ).
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6 Given that P and Q :
a write down the matrix PQ
b write down the matrix Q–1
Note : Although you are not required to be able to find the inverse of a given matrix, you may be
interested in a quick way to write down the inverse of a 2 2 matrix:
If A , then A1
A1
.
To obtain A–1
: a interchange the elements on the principal diagonal;
b reverse the signs of the elements on the secondary diagonal;
cdivide by the determinant of the original matrix.
Example
If A , then A (4) (6)
2
A–1 2
D7 The algebra of matricesAs a result of the definitions of the operations of matrices, the laws for the algebra of matrices
are mostly the same as the laws for the algebra of real numbers.
Note :
◗ This method does not work for 3 3 matrices or higher orders.
◗ If A 0, the matrix A has no inverse (because division by zero is not defined). Therefore,
the matrix is not invertible.
0.5 1.5
1 2
1 32 4
4 3
2 1
4 3
2 1
d b
c a a b
c d
2 4 4
3 6 1
7 4 1
1 2 2
1 3 1
3 2 0
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Hence, most algebraic operations with matrices are already quite familiar to us.
Examples
1 If 3A 4B 5C
then 3A 5C – 4B
A
13(5C – 4B)
2 (A B)(C D) AC AD BC BD
Remember :
◗ O, the zero matrix for matrix A, has been defined as the matrix with the same order as A
but having all its elements zeros.
◗ I, the identity matrix for matrix A, has been defined for square matrices only, being the
matrix having the same order as A, with all the elements on the principal diagonal being 1
and all the other elements being zero.
◗ A–1
, the inverse of matrix A, has been defined for square matrices only, being the matrix
such that AA–1 A
–1A I. The only matrices that have an inverse are square matrices
whose determinant 0.
119Appendix D Matrices and 3 3 Determinants
Name of law Real numbers Matrices
Commutative law for addition x y y x A B B A
Associative law for addition ( x y) z x ( y z) (A B) C A (B C)
Identity law for addition x 0 0 x x A O O A A
(0 is the identity element (O is the identity matrix forfor addition) addition, the zero matrix of A)
Identity law for multiplication x 1 1 x x A I I A A
(1 is the identity element (I is the identity matrix for
for multiplication) multiplication for matrix A)
Law of multiplicative inverse x x –1 x
–1 x 1 A A
–1 A
–1 A I
( x –1
is the multiplicative (A–1
is the multiplicative
inverse of x ) inverse of A)
Distributive law for C ( x y) Cx Cy k (A B) k A k B
multiplication ( x y)C xC yC (A B)k Ak Bk
x ( y z) xy xz A(B C) AB AC
( y z) x yx zx (B C)A BA CA
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3 (A B)(A I) A2 AI BA BI
A2 A BA B
4 A(A–1 I) AA
–1 AI
I A, or A I
5 A(A–1B) (AA–1)B IB B
6 AB A AB AI
A(B I), not A(B 1) because we cannot add a real number to a matrix
7 A–1
(A I) A–1
A A–1
I
I A–1
, or A–1 I
However, there is a difference between the algebra for matrices and the algebra for real numbers
when we are multiplying because, as we have already noted, in general:
A BB A.
(You are reminded that this statement does not mean that they can never be equal but that we
cannot assume they are equal, because usually they are not equal.)
Hence, care must be taken to maintain the correct order of matrices when multiplying.
For example, A(B C) (B C)A and ABA–1 A
–1AB (which would equal B).
The only occasion when multiplying matrices is commutative is when they are inverses of oneanother (AA
–1 A
–1A [ I]) or when one of them is the identity matrix (AI IA [ A]).
Example
Make K the subject of the equation AK B.
We proceed thus: AK B
A–1
(AK) A–1
B (not BA–1
)
(A
–1
A)K
A
–1
B K A
–1B
Note : We cannot add a matrix and a real number (e.g. A 2 makes no sense). But we can
always replace A by AI or by IA.
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Example
Solve the matrix equation AF 2F B, for F.
AF 2F B
AF 2IF B
(A 2I)F B (A 2I)
–1(A 2I)F (A 2I)
–1B
F (A 2I)–1
B
Exercises D7
1 Simplify, removing all brackets:
a A(A–1
B) b A(A–1
I)
c A(BA–1
) A(A–1
B) d AB(A–1
B B–1
A)e I
2f I
2 I
3
g (A I)2
h (A I)(B A)
i A(I – A–1
) I2
j A(A–1 I) – B
–1B – IA
2 Solve the following matrix equation for X:
a 2X – A B b C – 3X D
c 3(A – 2X) 2(3A – X) d X
3A
X –2
3B
3 Solve the following matrix equations for X:
a AX B b XA B
c 2A AX B d AX I B
e X–1 A f 3X – AX B
g AX X B h AX B C – X
i X XA B j 2X B C – 2XA
4 If AB AC:
a does it follow that B C (‘yes’ or ‘no’)?
b and BC, what is B equal to?
5 If AB
CA:a does it follow that B C (‘yes’ or ‘no’)?
b and BC, what is B equal to?
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Exercises D7 (continued)
6 Solve the following matrix equations for X:
a 3X 2XA C – B b A 2X B 3XC
D8 Expressing simultaneousequations in matrix form
Exercises D8
1 Find the following products and state their order:
a
b
2 Express as the product of two matrices:
a b c 3 Solve for x and y:
a
b
4 Express as a system of simultaneous equations:
a b
c d 4
5
6
x
y
z
1 2 3
0 1 2
1 2 0
6
7
8
3 x – 2 y 4 z
2 x 3 y – 5 z
x – 4 y 2 z
6
7
x
y
3 5
2 4
3
4
2 x 3 y
5 x – 2 y
29
27
x
y
13 1
12 1
8
9
2 x
3 y
3a – 5b 2c
2a 4b
5a – 7c
2 x – 3 y 4 z
x 2 y – 5 z
3 x – y 2 z
3 x 4 y
2 x – 5 y
x
y z
3 5 2
1 4 32 3 4
x
y
2 3
5 4
Note :
◗ The matrix, A B C (A B) C A (B C), the associative law, but we must
not change the order of the matrices when they are being multiplied.
◗ The matrices ABC, ACB, BCA, BAC, CAB and CBA are probably all different matrices.
◗ However, consideration will show you that if k is a constant (i.e. a real number), then
k (A B) (k A) B A (k B). Which matrix we multiply by k , either A B
or A or B, makes no difference to the final result. Although we must not change the
positions of the matrices , we may change the position of a constant, k .
◗ Note also that k A k AI A k I.
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5 Express each of the following systems of simultaneous equations as a single matrix equation:
a b
D9 Solving simultaneous linearequations using matricesA simple example is probably the best way to demonstrate how to solve simultaneous linear
equations using matrices.
Example
Solve the simultaneous equations
5x – 2 y 6
{ 3x – y 5
given that the inverse of is .
Steps
1 Express the equations as a single matrix equation:
2 Multiply both sides by the inverse matrix:
3 Multiply out the matrices on each side—we know the product on the left-hand side because:
A–1
A I .
x 4, y 7
Warning : The error most often made is in step 3. Remember that is not the same as
because for matrices A B B A.
This method involves knowing the inverse of the matrix formed by the coefficients in the
equation. You are not expected to be able to find the inverse of a given matrix, so either you
1 2
3 5
6
5
6
5
1 2
3 5
4
7
x
y
4
7
x
y
1 0
0 1
1 0
0 1
6
5
1 2
3 5
x
y
5 2
3 1
1 2
3 5
6
5
x
y
5 2
3 1
1 2
3 5
5 2
3 1
4 x – 5 y 6 z 9
7 x 3 y 5
3 x – 8 z 7
5a 7b – 3c 17
7a – 2b – 8c 13
3a 5b 5c 19
123Appendix D Matrices and 3 3 Determinants
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would be given the required inverse or you would be expected to deduce it by showing that the
product of two given matrices I.
Example
If M
and N
1
2
:
a find the matrix MN
b write down the matrix M–1
c express the given system of linear equations as a single matrix equation and use the
result of b above to solve these simultaneous equations:
x 2 y z –1
3x 5 y z 2{ 2x y 2z 9
Solutions
a MN
12
b M–1
12
Note : So that the matrix M will match the coefficients of the third equation, we write the third
equation as – 2x – y – 2z –9.
c M
M–1
M
1
2
12
x 3, y 1, z 2
3
–1
2
x
y
z
6
–2
–4
–1
2
–9
11 –5 –3
–8 4 2
–7 3 1
x
y
z
–1
2
–9
x
y
z
–1
2
–9
x
y
z
1 2 –1
3 5 –1
–2 –1 –2
11 –5 3
–8 4 2
–7 3 1
1 0 0
0 1 0
0 0 1
2 0 0
0 2 0
0 0 2
11 –5 –3
–8 4 2
–7 3 1
1 2 –1
3 5 –1
2 –1 –2
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Exercises D9
1 a Express the simultaneous equations
2 x – 3 y 9{5 x – 7 y 22
as a single matrix equation.
b Solve the above equations using matrices, given that the inverse of the matrix
is 2 You are given that if A then A
–1
A
1
a If P :
i evaluate P
ii write down the matrix P–1
b Use the result for P–1
above to solve the system of equations
3 x 2 y 4{5 x 4 y 10
3 If M and N :
a find the product MN
b write down the matrix M–1
c use the result of a above to solve the simultaneous equations below, showing each step
of the working:
4 x 3 y 7{ x – 2 y 10
4 The inverse of is . Use this result to solve the system
of equations .
5 If P and Q
12 :
a write down the matrix PQ
b write the system of equations below as a single matrix equation:
1 2 1
3 5 1
2 1 2
11 5 3
8 4 2
7 3 1
4a – b c 153a – 2b – c 13
a – b – c 3
1 2 3
2 5 7
1 3 5
4 1 1
3 2 1
1 1 1
2 3
1 –4
4 3
1 –2
3 2
5 4
d –b
–c a
a b
c d
–7 3
–5 2
2 3
5 7
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c use the result of a above to solve the simultaneous equations in b using matrices.
6 If P , Q , R :
a write down the matrices (i) PQ (ii) PR (iii) QR
b which two of these three matrices are inverses of each other?
c express the simultaneous equations
as a single matrix equation
d solve the above equations using matrices
7 Solve the following systems of simultaneous equations using matrices:
a given that the inverse of is b
given that if K
, then K
–1
c given that d given that
–1
e
given that
f given that I
g given that if M , then M–1
2 3 0
3 2 1
7 5 2
1 6 3
1 4 2
1 11 5
2a 2b –6
3a 2b c 0
7a 5b 2c –1
10 7 1
16 11 2
7 5 1
1 2 3
2 3 4
3 1 2
a 2b 3c –2
2a 3b 4c 0
3a b – 2c 17
1 0 0
0 1 00 0 1
23 5 –35
13 3 –2019 4 –29
7 5 5
3 2 55 3 4
23 x 5 y – 35 z –2
13 x 3 y – 20 z –119 x 4 y – 29 z –2
13 10 1
3 2 0
17 13 1
2 3 2
3 4 3
5 1 4
2 x 3 y 2 z 1
3 x 4 y 3 z 1
5 x y 4 z –1
1 0 0
0 1 0
0 0 1
5 3 0
13 7 1
16 9 1
2 3 3
3 5 5
5 3 4
2 p 3q 3r –2
3 p 5q 5r –4
5 p 3q 4r 0
3 7 2
0 3 1
5
2 0
2 4 1
5 10 3
15
29
9
2a – 4b – c 0
5a – 10b – 3c 1
15a – 29b – 9c
5
1 2 2
0 2 1
2 1 2
3 2 2
2 2 1
4 3 2
3 x 2 y – 2 z –3
2 x 2 y – z 1
–4 x – 3 y 2 z 0
a 2b – 2c 3
2a 5b – 4c 7
3a 7b – 5c 8
3
4 22 1 0
1 1 1
1 2
22 5 4
3 7 5
3
4 22 1 0
1 1 0
11 x – 5 y – 3 z –12
–8 x 4 y 2 z 10
7 x 3 y z 10
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h given that i
given that
–1
j given that I
k given that the inverse of is l
given that
–1
110
m given that
–1
17
n given that o given that
–1
110
Hence, matrices provide a general method for solving any number, n, of simultaneous equations
with n unknowns. A computer program can easily be designed for this purpose.
Note : Although a solution can be set out very concisely using the above formula, in an
examination a student should show every step of the solution.
By now you will have realised that the solution to any system of simultaneous equations is:
U C–1 K
where: U is the matrix formed by the unknowns;
C is the matrix formed by the coefficients;
K is the matrix formed by the constants.
–14 –4 106 –4 0
10 10 –10
2 3 23 2 3
5 5 4
2 x 3 y 2 z 13 x 2 y 3 z –6
5 x 5 y 4 z 0
10 0 0
0 10 0
0 0 10
35 5 –55
–215 –25 345
135 15 –215
4 5 7
7 2 5
3 3 4
4 p 5q 7r 21
7 p – 2q – 5r 0
3 p 3q 4r 13
19 26 8
17 24 9
5 5 1
3 2 6
4 3 5
5 5 2
3 x 2 y 6 z 1
4 x 3 y 5 z 1
5 x 5 y 2 z 0
35 25 1
30 20 05 5 1
2 3 2
3 4 35 5 5
2a 3b 2c 0
3a 4b 3c 15a 5b – 5c 55
36 1 22
3 0 2
31 1 19
2 3 2
5 2 6
3 5 3
2 x 3 y 2 z2
5 x 2 y 6 z 11
3 x 5 y 3 z4
3 2 2
5 3 3
7 5 4
3 2 0
41 26 1
46 29 1
3k – 2n 2t 0
5k – 3n 3t 1
7k 5n 4t 2
7 3 3
3 2 1
2
5 1
7 18 3
5 13 2
11
29 5
7 p – 18q 3r 5
5 p – 13q 2r 3
11 p – 29q
5r
8
1 0 0
0 1 0
0 0 1
7 5 7
3 2 4
2 3 10
32 71 6
38 84 7
5 11 1
32 x – 71 y – 6 z –1
38 x – 84 y – 7 z –1
5 x – 11 y – z 0
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D10 3 3 determinants: definitionand evaluation
is a square array of ‘elements’ having three rows and three columns. It is called a
‘3 3 determinant’ or a ‘determinant of third order’.
In Appendix C the value of a second order determinant was defined so that it provided a
shorthand method of solving two simultaneous equations in two unknowns. The value of a
third order determinant is defined so that it provides a shorthand method of solving three
simultaneous equations in three unknowns.
The value of the above determinant is defined as aei bfg cdh – gec – hfa – idb. There are
several ways of obtaining this result without having the very difficult task of committing it to
memory. We will use the method called the Rule of Sarrus. This method is the simplest but it
applies only to determinants of order three. (Later, when you study determinants of higher
orders, you will learn other methods and ones that are easier to program for a computer.)
The Rule of Sarrus
1 Write down the determinant, repeating the first two columns.
2 Obtain the products on the diagonals as shown below.
3 Add the lower products and add the upper products.
4 Subtract the sum of the upper products from the sum of the lower products.
Follow the application of this rule as we apply it to the general determinant
1
2, 3 (Sum gec hfa idb)
(Sum aei bfg cdh)
4 Value: (aei bfg cdh) – (gec hfa idb)
a b c a b
d e f d e
g h i g h
a b c
d e f g h i
a b c
d e f
g h i
128 Mathematics for Technicians
gec hfa idb
aei bfg cdh
a
d
g
b
e
h
c
f
i
a
d
g
b
e
h
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Example
Evaluate: Method: (Sum 36)
(Sum 13)
Value: (13) (36)
23
Exercises D101 Evaluate: ( Note: All the elements are exact numbers.)
a b c d e f g h
20 24 28
31 47 64
83 51 86
42 28 62
36 29 91
37 30 47
0 t n
t 0 x
n x 0
3 1 2
0 4 1
2 3 2
1 2 3
2 3 8
5 2 1
1 1 0
1 1 2
1 1 2
4 3 5
0 5 0
4 4 5
2 1 2
3 0 1
5 0 1
Hint : When copying down a determinant be very careful to include any negative signs. Check
your copy before working on it. If you copied it row by row, check it column by column. It is
very annoying to work on data that is later discovered to have been copied down incorrectly.
2 0 3
1 4 0
3 1 2
129Appendix D Matrices and 3 3 Determinants
–36 0 0
–16 0 3
2
–1
3
0
4
1
–3
0
–2
2
–1
3
0
4
1
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2 In each case below, evaluate the pronumeral:
a 9 b 10
c
5 d
3
e
D11 Solutions of simultaneous linear
equations using 3
3 determinantsA system of three linear equations in three unknowns may be solved algebraically by the
elimination method.
However, this method is usually very tedious and the equations are more easily solved using
determinants.
The general equations are: If we solve these equations by elimination we obtain the solution:
x
x , y
y, z
z
where: , which is the determinant formed by using the coefficients on the
left-hand side of the equation;
x
, which is the same determinant but with the coefficients of x
replaced by the constants (i.e. the numbers on the right-hand side);
y , which is again but with the coefficients of y replaced by the
constants;
z , which is again but with the coefficients of z replaced by the
constants.
a1 b1 d 1
a2 b2 d 2
a3 b3 d 3
a1 d 1 c1
a2 d 2 c2
a3 d 3 c3
d 1 b1 c1
d 2 b2 c2
d 3 b3 c3
a1 b1 c1
a2 b2 c2
a3 b3 c3
a1 x b1 y c1 z d 1
a2 x b2 y c2 z d 2
a3 x b3 y c3 z d 3
1 k 6
1 0 2
2 1 k
1 0 k
1 2 2
k 2 0
t t 1
4 1 0
2 t t
2 n 1
1 n 2
1 n 1
1 x 2
2 0 x
3 1 1
x 2 x 3 x
1 2 0
0 1 3
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Example
Given:
x
y
z Solution
x
x
1199 –1
y
y
31
89 2
z
z
5
1
7
9 3
(5) (62)
57
1 2 4
3 1 7
4 4 3
(80) (42)
38
1 4 3
3 7 2
4 3 3
(84) (65)
19
4 2 3
7 1 2
3 4 3
(17) 2
19
1 2 3
3 1 2
4 4 3
x 2 y 3z 4
3x y 2z 7
4x 4 y 3z 3
131Appendix D Matrices and 3 3 Determinants
12 8 –18
3 16 –36
1
3
4
–2
1
–4
3
–2
3
1
3
4
–2
1
–4
(2)
(–17)
–9 32 42
12 12 84
4 –
1
–4
3 4
1
(65)
(84)
–84 6 36
–21 –32 –27
1
3
4
4
–7
–3
3
–2
3
1
3
4
4
–7
–3
(–42)
(–80)
16 28 18
–3 56 –48
1
3
4
–2
1
–4
1
3
4
(62)
(5)
4
–7
–3
–2
1
–4
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Exercises D11
1 Solve for the pronumerals using determinants (do not use a calculator):
a b 2 p – 4q 3r 8 0
p 3q – 2r 2 0
3 p – 5q – 4r – 4 0
2 x y 3 z 4
3 x – y – 4 z 5
4 x 3 y 2 z 1
Hints : a Before using the Rule of Sarrus to solve a set of simultaneous equations, make sure
that:
i all the equations have the pronumerals in the same order ;
ii all the pronumerals are on the left-hand sides of the equations and all the
constants are on the right-hand sides;
iii if a pronumeral is absent from an equation, write it in with a zero coefficient—for
example, if there is no y in an equation, write it in as 0 y .
b When copying down an array to work on, be careful to place the elements in a neat
rectangle so that the diagonal elements are approximately in straight lines. An untidy
array leads to errors in computation.
c Before working on your arrays, check that you have made no errors, especially by
omitting any negative signs. Discover any errors before you start to multiply.
d When multiplying out diagonals, either mentally or with a calculator, ignore any
negative signs present. Decide after the multiplication whether the product should be
positive or negative.
Points to note
• After much practice, the value of a determinant can be found on a calculator, using the Rule
of Sarrus , showing no intermediate results. The diagonal products can be summed using the
M
and M–
keys. However, for the time being, you are advised to write down the product of
each diagonal before adding it to the calculator memory, as in the examples given. This
allows you to check your work more easily and also allows for credit to be given in anexamination for knowledge of the method even if an error is made during the computation.
• When you have solved a set of simultaneous equations, check your result by substituting
your values back into the original equations. You will be surprised how often careless errors
are made during a long series of calculations.
• When using a calculator, always work with all the significant figures in the data. State your
result to the appropriate number of significant figures, that is, the number required or
justified.
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c d 2 Solve for the pronumerals using a calculator:
a b
c 3
The equations for equilibrium of this beam are:
Vertical forces: 0.6F 2 – 2 0.8F 3 0
Horizontal forces: F 1 0.8F 2 – 0.6F 3 0
Moments about point P: 0.6F 2 – 4 2.4F 3 0
Using determinants, solve for F 1, F 2 and F 3, stating the results correct to 3 significant
figures. Explain the meaning of any negative values.
4
2(3.7k – 1.4t ) 3(4.3t 2.6w) 8.7 0
4(2.3k 1.7w) – 5.3w – 6.1 0
3(3.9k 4.2t ) 0
Hint : In b, multiply or divide both sides of each equation by a constant so as to obtain simpler
numbers.
State results correct to 3 significant figures.
0.002 x 0.003 y 0.001 z 0.010.3 x – 0.4 y 0.2 z 0.8
4000 x 5000 y – 3000 z 4000
37 x 49 y 76 z 9868 y – 34 x – 93 z 136
82 z – 36 y 29 x –72
4n – 3 p 5t –3
3n 4 p –6
5 p – 4t –4
3 x 2 y 4 z –7
2 y – 3 z 4 x 6
4 z – 5 x – 2 y –7
133Appendix D Matrices and 3 3 Determinants
1 m 1 m 1 m
2 kN
F
P
2 F 3
F 1
37° 53°
12 V I 1 I 2 I 3
6 V
12 V
2 3
4
4
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Applying Kirchhoff’s laws to this network:
4 I 1 2( I 1 – I 2) 12
2( I 2 – I 1) 3( I 2 – I 3) 6
3( I 3 – I 2) 4 I 3 12
Use determinants to solve for I 1, I 2 and I 3, stating their values correct to 3 significant
figures.5 A developer receives council permission to divide his land of area 44 ha into 100 blocks,
the only areas allowed for a block being 2 ha, 0.5 ha and 0.2 ha. He prices the 2 ha blocks
at $100 000 each, the 0.5 ha blocks at $40 000 each and the 0.2 ha blocks at $20 000 each.
He sells all the blocks, the total gross income from the sales being $3 200 000. How many
blocks of each size did he sell?
6 A firm has a stock of three different bronze alloys. Alloy A consists of 95% copper, 3% tin
and 2% zinc. Alloy B consists of 90% copper, 9% tin and 1% zinc. Alloy C consists of 80%
copper, 15% tin and 5% zinc.
How many kilograms of each of these alloys must be melted and mixed in order to produce
100 kg of a new alloy that consists of 87% copper, 9.6% tin and 3.4% zinc?