Introduction
Matrix theory is a powerful mathematical tool for dealing with data as a whole
rather than the individual items of data. Matrices are especially useful in the
theory of equations. They can be used to solve systems of simultaneous linear
equations. Determinants are related to matrices and are useful for determining
whether or not a unique solution exists. In some cases using determinants, the
solution for each unknown can be expressed explicitly in terms of the coefficients
of the equations by applying what is known as Cramer’s rule. Systems
of simultaneous linear equations occur, for example, when optimizing a function
using Lagrange multipliers or when trying to find the equilibrium prices
of interdependent commodities. As we shall see, matrices can be added and
in some cases multiplied together. In economics, business, and finance, many
basic theoretical models are linear in that they are described in some way by
linear functions. Analyzing these models is made simpler by matrix algebra.
10.2 Matrix Operations
A rectangular array of mn numbers in m rows and n columns is called a matrix
of size m×n (‘m by n’). The array is enclosed in square or, sometimes, curved
brackets. The (i, j)-entry of a matrix M is the entry in the ith row and jth
209
210 Elements of Mathematics for Economics and Finance
column of M. This entry can be denoted simply by Mij . For example:
M =
3 −1 0
0 2 −5
is a 2 × 3 matrix
and
N =
5 1
2 −1
is a 2 × 2 matrix.
For the matrix M, we have M11 = 3, M13 = 0, M23 = −5, and so on.
An n × n matrix, that is one that has as many rows as columns, is called a
square matrix. The matrix N is square. A 1×n matrix is called a row matrix
or row vector of length n. An n × 1 matrix is called a column matrix or
column vector of height n. For example:
4
1
is a column matrix of height 2
and
0 1 −3
is a row matrix of length 3.
The transpose of an m×n matrix M is the n×m matrix whose ith row is
the ith column of M (i = 1, 2, . . . n). The matrix is denoted by Mt and called
‘M transpose’. Another way to define Mt is as the n × m matrix whose (i, j)-
entry is Mji. It is immediately clear that the transpose of Mt is M. That is,
(Mt)t = M. For example,
3 5 −1
t =
⎡
⎣
3
5
−1
⎤
⎦ ,
4 3
0 −2
t
4 0
3 −2
,
1 4 −1
2 0 −3
t
⎡
⎣
1 2
4 0
−1 −3
⎤
⎦.
A matrix M that is its own transpose, so that Mt = M, is said to be
symmetric. Obviously, only square matrices can be symmetric.
⎡
⎣
3 1 0
1 −2 6
0 6 2
⎤
⎦is a symmetric matrix.
We shall introduce some operations that can be performed on matrices.
The three basic ones are: scalar multiplication, matrix addition, and matrix
multiplication.
- Matrices and Determinants 211
10.2.1 Scalar Multiplication
The term ‘scalar’ in this context simply means a number as opposed to a matrix.
The reasons for use of this term are historical. It is still a useful term if we wish
to distinguish scalar from matrix multiplication of matrices.
If λ is any number and M any matrix, the scalar multiple of M by λ is
the matrix, denoted by λM, obtained by multiplying each entry of M by λ, so
the (i, j)-entry of λM is λMij . Obviously, λM and M have the same size. The
following are some examples:
36 −16
0 24
= 4
9 −4
0 6
,
−2
4 1 −2
−3 0 1
=
−8 −2 4
6 0 −2
.
We write −M rather than (−1)M. So, for instance
−
1 −3
−4 0
=
−1 3
4 0
.
Suppose that two firms A and B each produce two goods G1, G2. A consumer
is supplied with both goods by both firms. The quantities supplied over
a particular period can be represented by a matrix
Q =
Q11 Q12
Q21 Q22
,
which we shall call the supply matrix, where the first row gives the quantity
supplied of goods G1 and G2, respectively, by firm A and the second row that
by firm B. If the consumer increases all the quantities bought from each firm
by 20%, the new supply matrix would be (1.2)Q.
In theoretical discussions, it is sometimes useful to distinguish scalars from
matrices by denoting scalars by lowercase Greek letters and matrices by uppercase
latin letters. For instance, the following easy to see matrix rule
λ (μM) = (λμ)M
says that multiplying a matrix M by a scalar μ and then by a scalar λ is the
same as multiplying M by λμ. For example, 3 (5M) = 15M.
212 Elements of Mathematics for Economics and Finance
10.2.2 Matrix Addition
Matrices of the same size can be added. If M and N are matrices of the same
size, then their sum M +N is the matrix whose (i, j)-entry is Mij +Nij. That
is M + N is obtained by adding corresponding entries of M and N. Clearly
M + N will have the same size as M and N. For example,
5 −4
3 2
+
−2 4
1 −1
=
3 0
4 1
,
⎡
⎣
5
−2
1
⎤
⎦ +
⎡
⎣
3
−4
−2
⎤
⎦ =
⎡
⎣
8
−6
−1
⎤
⎦ .
Suppose
Q =
10 35
15 18
, and Q
=
25 10
40 5
are the supply matrices (see Section 10.2.1) for two successive periods of a year.
Then their sum
Q + Q
=
35 45
55 23
is the supply matrix for the combined two-year period.
We can also define a matrix M − N in an obvious way by subtracting
corresponding entries:
5 −4
3 2
−
−2 4
1 −1
=
7 −8
2 3
.
It is clear that M − N = M + (−N).
Matrix addition is commutative. This is the formal way of saying that the
order in which addition is performed is unimportant. That is, M+N = N +M
for any two matrices M, N of the same size.
The m × n zero matrix is the m × n matrix with all zero entries. It is
denoted simply by 0, the size m×n being clear usually from the context. Clearly
M +0 = 0+M = M for any matrix M. We have also M −M = 0 = −M +M.
10.2.3 Matrix Multiplication
If A =
a1 a2 . . . am
is a 1 × m row matrix and B =
⎡
⎢⎢⎢⎣
b1
b2
…
bm
⎤
⎥⎥⎥⎦
is an
m × 1 column matrix , then we define the product AB to be the number
a1b1 + a2b2 + . . . + ambm.
- Matrices and Determinants 213
More generally, if M, N are matrices, then we can define their product MN
if the number of columns in M is the number of rows in N; say M is m × p
and N is p×n. Then MN is the m×n matrix whose (i, j)-entry is the number
MiNj, where Mi is the ith row of M and Nj the jth column of N. The following
example illustrates this operation for some particular matrices.
Example 10.1 - If M =
3 4 −1
and N =
⎡
⎣
2
1
7
⎤
⎦, then
MN = 3× 2 + 4 × 1 + (−1) × 7 = 6+4 − 7 = 3.
If M =
2 −3 1
and N =
⎡
⎣
4
1
−5
⎤
⎦, then
MN = 2× 4 + (−3) × 1 + 1 × (−5) = 8 − 3 − 5 = 0.
- If M =
⎡
⎣
2 −1 4
1 0 2
2 3 −8
⎤
⎦ and N =
⎡
⎣
5 2
4 −3
1 0
⎤
⎦, then
MN =
⎡
⎣
10 7
7 2
14 −5
⎤
⎦ ,
since
M1N1 = 2× 5 + (−1) × 4 + 4 × 1 = 10,
M1N2 = 2× 2 + (−1) × (−3) + 4 × 0 = 7,
M2N1 = 1× 5 + 0 × 4 + 2 × 1 = 7,
M2N2 = 1× 2 + 0 × (−3) + 2 × 0 = 2,
M3N1 = 2× 5 + 3 × 4 + (−8) × 1 = 14,
M3N2 = 2× 2 + 3 × (−3) + (−8) ×0 = −5.
214 Elements of Mathematics for Economics and Finance - If A =
2 4
1 0
and B =
1 −1
2 1
, then
AB =
2 4
1 0
1 −1
2 1
=
2 × 1 + 4 ×2 2× (−1) + 4 × 1
1 × 1 + 0 ×2 1× (−1) + 0 × 1
=
10 2
1 −1
and
BA =
1 −1
2 1
2 4
1 0
=
1 × 2 + (−1) ×1 1× 4 + (−1) × 0
2 × 2 + 1 ×1 2× 4 + 1 × 0
=
1 4
5 8
.
- If A =
5 1
and B =
3 1 2
−2 0 4
, then
AB =
5 × 3 + 1 × (−2) 5 × 1 + 1 ×0 5× 2 + 1 × 4
13 5 14
.
- If A =
⎡
⎣
3 1 −4
0 2 1
5 −2 −3
⎤
⎦ and B =
⎡
⎣
2
−1
1
⎤
⎦, then
AB =
⎡
⎣
3 1 −4
0 2 1
5 −2 −3
⎤
⎦
⎡
⎣
2
−1
1
⎤
⎦
⎡
⎣
3 × 2 + 1 × (−1) + (−4) × 1
0 × 2 + 2 × (−1) + 1 × 1
5 × 2 + (−2) × (−1) + (−3) × 1
⎤
⎦
⎡
⎣
1
−1
9
⎤
⎦ .
- Matrices and Determinants 215
- If A =
1 1
−1 −1
and B =
1 −1
−1 1
, then
AB =
1 1
−1 −1
1 −1
−1 1
=
1 × 1 + 1 × (−1) 1 × (−1) + 1 × 1
(−1) × 1 + (−1) × (−1) (−1) × (−1) + (−1) × 1
=
0 0
0 0
.
Example 10.2
Two firms A and B each produce three goods G1, G2, and G3. The prices
per unit for each good from the two firms are represented by the matrix P =
PA1 PA2 PA3
PB1 PB2 PB3
, where the first row gives the prices per unit for goods
G1, G2, G3, respectively, supplied by firm A and the second row gives those
supplied by firm B.
A consumer wishes to buy quantities Q1, Q2, Q3, respectively, from one of
the firms. The matrix Q =
⎡
⎣
Q1
Q2
Q3
⎤
⎦ represents the quantities. Then PQ is the
2×1 column matrix
PA1Q1 + PA2Q2 + PA3Q3
PB1Q1 + PB2Q2 + PB3Q3
, where the top entry in the
column is the cost of buying the goods from firm A and the bottom entry the
cost from firm B.
Notes
- To get the first row entries of the product MN of two matrices M and N,
multiply the first row of M in turn by each column of N. Do this for the
second row and so on. This constructs MN row by row. So for instance, in
Example 10.1.2,
2 −1 4
⎡
⎣
5
4
1
⎤
⎦ = 10 − 4 + 4 = 10
and
2 −1 4
⎡
⎣
2
−3
0
⎤
⎦ = 4+(−1)(−3) + 0 = 7
are the entries of the first row of MN.
216 Elements of Mathematics for Economics and Finance
- If M and N are matrices, it is easy to see that the matrix products MN
and NM are both defined when and only when M, N are square matrices
of the same size. However, MN and NM are not in general the same.
See Example 10.1.3 for instance where AB = BA. Thus, unlike matrix
addition, matrix multiplication is not commutative.
Note that in Example 10.1.2, we cannot define the product NM.
In Example 10.1.3, the matrices A, B can be multiplied in two ways to give
the products AB and BA. To specify AB, for instance, we can say this is
the product of A multiplied by B on the right or B multiplied on the left
by A.
The diagonal of a matrix M consists of all the entries of the form Mii. For
instance, the diagonal entries of the matrix
M =
⎡
⎣
3 4 1 1
0 1 2 0
0 0 −5 1
⎤
⎦
are 3, 1, −5. For any matrix M, it is easy to see that Mt and M will have the
same diagonal.
The square n×n matrix in which every diagonal entry is 1 and every entry
off the diagonal is 0 is called the identity n×n matrix. This matrix is denoted
by In; or simply by I if its size is clear from the context.
I2 =
1 0
0 1
, I3 =
⎡
⎣
1 0 0
0 1 0
0 0 1
⎤
⎦.
Identity matrices act like ‘ones’ in the sense that AIn = ImA = A if A is any
m × n matrix.
A useful matrix rule is
(λM)N = M (λN) = λ (MN) .
Essentially, this means that when multiplying matrices together, any scalar
factor, such as λ, can be taken ‘outside’ the multiplication process. For example,
if A and B are as in Example 10.1.3, then
1
3A
(6B) =
1
3
(6)AB = 2AB =
20 4
2 −2
.
Two more rules, known as the distributive laws, allows us to ‘open’ brackets:
A(B + C) = AB + AC
- Matrices and Determinants 217
and
(A + B)C = AC + BC.
Conversely, the rules can be regarded as one of factorization. For instance,
if A, B, C are matrices then:
(a) 2AB − AC = A(2B − C);
(b) BA − A = (B − I)A.
(Here, I is the m × m identity matrix, where m is the number of rows in A.)
This is true because ImA = A. (Why would it be wrong to write B −1 instead
of B − I?)
If M is a square m × m matrix, we can define powers of M: M2 = MM,
M3 = M2M = MMM, and so on. In general, for any integer n ≥ 1, we have
Mn = Mn−1M, where we take M0 = Im, the identity m × m matrix. It is
easy to see that Mn is just M multiplied by itself n times. This defines the
nth power Mn of a square matrix for any integer n ≥ 0, similar to the way in
which xn is defined for a number x.
Example 10.3
If A is the matrix defined by
A =
1 3
2 0
compute A2 and A3.
Solution.
A2 =
1 3
2 0
1 3
2 0
=
7 3
2 6
A3 = AA2 =
1 3
2 0
7 3
2 6
=
13 21
14 6
The question that naturally arises is whether this analogy of powers of
matrices with powers of numbers extends to negative integer exponents. In
particular, can we assign any meaning to M−1? It turns out that we can define
M−1 in certain cases and we can test whether or not M−1 exists.
We shall say that a square matrix M is invertible if there is a matrix N
such that MN = NM = I. If N exists, we write M−1 for N. We call M−1
the inverse matrix of M. (For short, we can say ‘M inverse’ for the matrix
M−1.) It is easy to see that
M−1
−1 = M.
218 Elements of Mathematics for Economics and Finance
It is important to note that not all square matrices are invertible. To determine
whether or not a matrix is invertible, we need the concept of the
determinant of a square matrix.
First we consider a 2 × 2 matrix
M =
a b
c d
.
The determinant of M denoted by |M| or
a b
c d
,
is the number ad − bc. For example:
2 −1
3 5
= 10 − (−3) = 13,
6 3
4 2
= 12 − 12 = 0.
It can be shown that a 2 × 2 matrix is invertible if, and only if, its determinant
is not zero. If M is invertible, its inverse is obtained in the following
way.
If M =
a b
c d
and |M| = ad − bc = 0, then M−1 =
1
ad − bc
d −b
−c a
The matrix M−1 is therefore a scalar multiple of the matrix
d −b
−c a
;
the scalar being 1/|M|. To illustrate this, perform the matrix multiplication
1
M
d −b
−c a
a b
c d
=
1
ad − bc
ad − bc 0
0 ad − bc
=
1 0
0 1
= I.
For example,
6 3
4 2
= 0,
- Matrices and Determinants 219
so the matrix
6 3
4 2
is not invertible.
Since
3 5
2 4
= 12 − 10 = 2,
the matrix
M =
3 5
2 4
is invertible and
M
−1 =
1
2
4 −5
−2 3
.
Example 10.4
Given that
A =
1 −2
3 4
and B =
4 1
0 −3
,
determine the 2 × 2 matrices X and Y satisfying
- AX = B;
- Y A = B.
Solution. Since |A| = 4− (−6) = 10 = 0, then A−1 exists and
A
−1 =
1
10
4 2
−3 1
.
- Multiply both sides of the equation AX = B on the left by A−1 to get
A−1AX = A−1B. That is, IX = A−1B and so X = A−1B. Therefore,
X =
1
10
4 2
−3 1
4 1
0 −3
=
1
10
16 −2
−12 −6
.
(Here we used the fact that A−1A = I, the identity 2×2 matrix, and that
IX = X.)
- Multiply both sides of the equation on the right by A−1 to get Y AA−1 =
BA−1, which gives Y = BA−1. Therefore,
Y =
1
10
4 1
0 −3
4 2
−3 1
=
1
10
13 9
9 −3
.
220 Elements of Mathematics for Economics and Finance
10.3 Solutions of Linear Systems of Equations
Two simultaneous equations in two unknowns can be solved uniquely if the
matrix of coefficients is invertible. Specifically, if
ax + by = c
px + qy = r
are the equations in the unknowns x, y, the matrix of coefficients is
M =
a b
p q
.
Then
M
x
y
=
ax + by
px + qy
=
c
r
so that the two given simultaneous equations are equivalent to one matrix
equation:
M
x
y
=
c
r
.
If M is invertible, then multiplying both sides on the left by M−1 gives the
unique solution
x
y
= M
−1
c
r
.
Example 10.5
Solve the linear system of equations
7x + 3y = 41
3x + 2y = 19
(10.1)
Solution. The matrix form of (10.1) is
7 3
3 2
x
y
=
41
19
.
Since
7 3
3 2
= 14 − 9 = 5 = 0,
- Matrices and Determinants 221
the matrix of coefficientsM =
7 3
3 2
is invertible andM−1 =
1
5
2 −3
−3 7
.
The solution is
x
y
= M
−1
41
19
=
1
5
2 −3
−3 7
41
19
=
1
5
25
10
.
So x = 25
5 = 5 and y = 10
5 = 2.
Example 10.6
The demand and supply equations for a good are given by P +4QD = 70 and
P − QS = 5. Determine the equilibrium price and quantity.
Solution. To find the equilibrium price P and quantity Q = QD = QS, we
solve the equations
P + 4Q = 70
P − Q = 5.
In matrix form, this is equivalent to
1 4
1 −1
P
Q
=
70
5
.
There is a unique solution:
P
Q
=
1 4
1 −1
−1
70
5
= −1
5
−1 −4
−1 1
70
5
= −1
5
−90
−65
=
18
13
.
The equilibrium values are therefore P = 18, Q = 13.
222 Elements of Mathematics for Economics and Finance
10.4 Cramer’s Rule
This simple rule allows us to express the solution of two simultaneous equations
in two unknowns explicitly, assuming there is a unique solution. Given the
equations
ax + by = c
px + qy = r
then Cramer’s rule states that if
a b
p q
= 0 (equivalently, if the matrix of
coefficients is invertible), then
x =
c b
r q
a b
p q
and y =
a c
p r
a b
p q
.
Observe that the determinant
c b
r q
in the equation for x is obtained by
replacing the column of coefficients of x in the determinant of the matrix of
coefficients by the column of constants on the right-hand side of the equation;
and similarly for y.
Example 10.7
Solve the simultaneous equations
6x + 7y = 10
4x + 5y = 8
using Cramer’s rule.
Solution. Using Cramer’s rule, we have
x =
10 7
8 5
6 7
4 5
50 − 56
30 − 28
= −3
and
y =
6 10
4 8
6 7
4 5
48 − 40
2
= 4.
- Matrices and Determinants 223
(In geometric terms, x and y are the coordinates of the point of intersection of
the two lines with equations 6x + 7y = 10 and 4x + 5y = 8.)
10.5 More Determinants
In Section 10.2, we introduced for 2 × 2 matrices the idea of a determinant.
Now we consider 3 × 3 matrices. The determinant of a 3 × 3 matrix
M =
⎡
⎣
a b c
d e f
g h i
⎤
⎦
is the number
a
e f
h i
− b
d f
g i
- c
d e
g h
which we denote by |M| or
a b c
d e f
g h i
.
Example 10.8
Evaluate the determinant
2 3 5
3 1 2
1 4 3
.
Solution. Using the definition of a determinant of a 3 × 3 matrix
2 3 5
3 1 2
1 4 3
=
2
1 2
4 3
− 3
3 2
1 3 - 5
3 1
1 4
= 2(3− 8) − 3(9 − 2) + 5(12 − 1)
= −10 − 21 + 55
= 24.
As in the 2 × 2 case, it is true that a 3 × 3 matrix M is invertible if, and
only if, its determinant |M| = 0; but it is not as easy to describe M−1 in the
3 × 3 case. A method for constructing M−1 (when it exists) is known as the
adjoint method. To describe this, we need the concept of a cofactor.
224 Elements of Mathematics for Economics and Finance
The (i, j)-cofactor of a 3×3 matrix M is the determinant of the 2×2 matrix
obtained by deleting the ith row and jth column of M, multiplied by (−1)i+j .
Note that (−1)i+j is +1 or −1 according to whether i + j is even or odd,
respectively. The pattern - − +
− + − - − +
gives the (i, j) positions corresponding to +1 and −1. For example, the (3, 1)-
cofactor of the matrix
M =
⎡
⎣
3 2 −4
1 3 2
4 1 −1
⎤
⎦
is
+
2 −4
3 2
= 4+12 = 16.
Similarly, the (2, 3)-cofactor is
−
3 2
4 1
= −(3 − 8) = 5.
Note The value of the (i, j)-cofactor does not depend on the value of the
(i, j)−entry.
The cofactor matrix of a 3×3 matrix M is the 3×3 matrix whose (i, j)-
entry is the (i, j)-cofactor of M. For example, the cofactor matrix of the matrix
M in the previous example is
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎣
3 2
1 −1
−
1 2
4 −1
1 3
4 1
−
2 −4
1 −1
3 −4
4 −1
−
3 2
4 1
2 −4
3 2
−
3 −4
1 2
3 2
1 3
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎣
−5 9 −11
−2 13 5
16 −10 7
⎤
⎦.
The transpose of the cofactor matrix of a 3 × 3 matrix M is known as the
adjoint matrix, denoted by adjM.
The following statement describes how M−1 may be computed by the
method known as the adjoint method.
If M is a 3 × 3 matrix and |M| = 0, then M−1 =
1
|M| adjM.
- Matrices and Determinants 225
Thus M−1 is a scalar multiple of adjM; the scalar being the number 1
|M| .
Example 10.9
Find the inverse of the matrix
M =
⎡
⎣
3 2 −4
1 3 2
4 1 −1
⎤
⎦.
Solution. The determinant of M is
3 2 −4
1 3 2
4 1 −1
= 3
3 2
1 −1
− 2
1 2
4 −1
- (−4)
1 3
4 1
= 3(−5) − 2(−9) − 4(−11)
= −15 + 18 + 44
= 47.
Since |M| = 47 = 0, then M−1 exists and M−1 = 1
|M| adjM, where adjM is
the transpose of the cofactor matrix of M (see above). Therefore
M
−1 =
1
47
⎡
⎣
−5 −2 16
9 13 −10
−11 5 7
⎤
⎦.
If we look again at the definition of the determinant of a 3 × 3 matrix M,
we see that |M|, the determinant of M, is obtained by multiplying each entry
in the first row of M by the corresponding cofactor and adding.
An interesting fact is that there is nothing special about the first row of M.
Multiplying each term in any row (or column) by the corresponding cofactor
and adding gives the same number; namely |M|. (This is a useful check that
the cofactor matrix has been computed correctly.)
For example, for the matrix
M =
⎡
⎣
3 2 −4
1 3 2
4 1 −1
⎤
⎦,
the cofactor matrix is ⎡
⎣
−5 9 −11
−2 13 5
16 −10 7
⎤
⎦.
226 Elements of Mathematics for Economics and Finance
The determinant of M is |M| = 3(−5)+2(9)+(−4)(−11) = −15+18+44 = 47.
This is the expansion of the determinant of M by its first row. Expanding
by, say, the third column gives (−4)(−11) +2(5) + (−1)(7) = 44 +10−7 = 47
again.
Incidentally, if we expand a row using the cofactor of another row, we will
always get 0. For example, expanding along the first row of M using the cofactors
of the third row gives 3(16)+2(−10)+(−4)(7) = 48−20−28 = 0. These
expansion properties of determinants form the basis of the adjoint method for
finding inverses and of Cramer’s rule.
Example 10.10
Determine the inverse of the matrix M, where
M =
⎡
⎣
1 −3 −2
4 1 2
0 6 5
⎤
⎦ .
Solution. The cofactor matrix of M is
C =
⎡
⎣
−7 −20 24
3 5 −6
−4 −10 13
⎤
⎦ .
The determinant is
|M| = 1(−7) − (−3)(20) + (−2)(24) = −7 + 60 − 48 = 5.
Therefore
M
−1 =
1
|M| adjM =
1
5Ct =
1
5
⎡
⎣
−7 3 −4
−20 5 −10
24 −6 13
⎤
⎦. (10.2)
Solving a system of two simultaneous equations in two unknowns using inverse
matrices or Cramer’s rule extends naturally to the case of three equations
in three unknowns. If the matrix
M =
⎡
⎣
a1 a2 a3
b1 b2 b3
c1 c2 c3
⎤
⎦
- Matrices and Determinants 227
of coefficients of a system of three equations
a1x + a2y + a3z = p
b1x + b2y + b3z = q
c1x + c2y + c3z = r
has non-zero determinant and so is invertible, then the simultaneous equations
have a unique solution given by:
⎡
⎣
x
y
z
⎤
⎦ = M
−1
⎡
⎣
p
q
r
⎤
⎦.
Alternatively, we can use Cramer’s rule, extended from Section 10.4 in an
obvious way to three equations in three unknowns. This gives
x =
|Mx|
|M|, y=
|My|
|M|, z=
|Mz|
|M| ,
where Mx is the matrix obtained by replacing the column
⎡
⎣
a1
b1
c1
⎤
⎦ of coefficients
of x in M by the column of constants
⎡
⎣
p
q
r
⎤
⎦. Similarly for My and Mz.
Example 10.11
Solve the system of three simultaneous equations
x − 3y − 2z = 5,
4x + y + 2z = 116,
6y + 5z = 47.
Solution. In matrix form, the system can be written as one equation:
M
⎡
⎣
x
y
z
⎤
⎦ =
⎡
⎣
5
116
47
⎤
⎦ ,
where
M =
⎡
⎣
1 −3 −2
4 1 2
0 6 5
⎤
⎦
228 Elements of Mathematics for Economics and Finance
is the matrix of coefficients.
From Example 10.10, we know that M is invertible and M−1 is given in
(10.2). Therefore
⎡
⎣
x
y
z
⎤
⎦ = M
−1
⎡
⎣
5
116
47
⎤
⎦ =
1
5
⎡
⎣
−7 3 −4
−20 5 −10
24 −6 13
⎤
⎦
⎡
⎣
5
116
47
⎤
⎦
1
5
⎡
⎣
−35 + 348 − 188
−100 + 580 − 470
120 − 696 + 611
⎤
⎦ =
1
5
⎡
⎣
125
10
35
⎤
⎦ =
⎡
⎣
25
2
7
⎤
⎦.
Therefore x = 25, y = 2, and z = 7.
Alternatively, we can use Cramer’s rule. Here,
Mx =
⎡
⎣
5 −3 −2
116 1 2
47 6 5
⎤
⎦,
so
|Mx| = 5(5 − 12) − (−3)(580 − 94) − 2(696 − 47) = 125.
Therefore
x =
|Mx|
|M| =
125
5
= 25,
and then y, z are obtained in a similar way, noting that
My =
⎡
⎣
1 5 −2
4 116 2
0 47 5
⎤
⎦
and
Mz =
⎡
⎣
1 −3 5
4 1 116
0 6 47
⎤
⎦.
Example 10.12
A consumer’s utility function is U(x, y) = xy + x + 2y, where x is the number
of units of good Gx and y the number of units of good Gy. The price per unit
of Gx is 2 (units of money), and the price per unit of Gy is 5. What is the
maximum utility if the consumer’s budget is 91?
- Matrices and Determinants 229
Solution. The objective function is U and the constraint is 2x + 5y = 91.
Form the Lagrangian function
F = xy + x + 2y + λ(91 − 2x − 5y).
The stationary points of F occur where
0 = Fx = y + 1 − 2λ,
0 = Fy = x + 2 − 5λ
and 0 = Fλ = 91 − 2x − 5y.
We therefore have the following three equations in the three unknowns x,
y, and λ:
−y + 2λ = 1,
x − 5λ = −2,
2x + 5y = 91.
We can easily solve these as before by eliminating λ between the first two
equations and then using the constraint equation to solve for x and y.
Another method is to use matrix inversion. The matrix of coefficients is
M =
⎡
⎣
0 −1 2
1 0 −5
2 5 0
⎤
⎦ ,
so that
M
⎡
⎣
x
y
λ
⎤
⎦ =
⎡
⎣
1
−2
91
⎤
⎦.
Using the adjoint method, we can compute
M
−1 =
1
20
⎡
⎣
25 10 5
−10 −4 2
5 −2 1
⎤
⎦.
Then
⎡
⎣
x
y
λ
⎤
⎦ = M
−1
⎡
⎣
1
−2
91
⎤
⎦ =
1
20
⎡
⎣
25 − 20 + 455
−10 + 8 + 182
5 + 4 + 91
⎤
⎦ =
1
20
⎡
⎣
460
180
100
⎤
⎦ .
Therefore, x = 23, y = 9, and λ = 5. So the maximum utility is U(23, 9) = 248.
Cramer’s rule could also have been used to solve the previous problem. We
use it for the next problem.
230 Elements of Mathematics for Economics and Finance
Example 10.13
Find the equilibrium prices of three interdependent commodities whose prices
P1, P2, P3 satisfy:
P1 + P2 + 3P3 = 37,
3P1 + 2P2 + 4P3 = 79,
2P1 + 3P2 + 5P3 = 76.
Solution. The determinant of the matrix of coefficients is
1 1 3
3 2 4
2 3 5
= (10 − 12) − (15 − 8) + 3(9 − 4) = 6.
Therefore by Cramer’s rule
P1 =
37 1 3
79 2 4
76 3 5
÷ 6 = 90 ÷ 6 = 15,
P2 =
1 37 3
3 79 4
2 76 5
÷ 6 = 42 ÷ 6 = 7,
P3 =
1 1 37
3 2 79
2 3 76
÷ 6 = 30 ÷ 6 = 5.
10.6 Special Cases
The solutions of systems of simultaneous equations considered in this chapter
have been for the case when there are as many unknowns as equations and
also the matrix of coefficients is invertible (or, equivalently, has non-zero determinant).
In this case, the solution is unique. What happens if the matrix of
coefficients is not invertible; that is, if its determinant is zero? In this case, the
system is either inconsistent and has no simultaneous solution or else it has
infinitely many solutions. - Matrices and Determinants 231
To illustrate this point with a very simple example, consider the two systems
of equations
(a) x + 3y = 4 (b) x + 3y = 5
2x + 6y = 8 2x + 6y = 9
In both cases, the matrix of coefficients is
1 3
2 6
and its determinant is
1 3
2 6
= 0.
In case (a), the second equation is twice the first equation. (The second
equation is therefore redundant.) Therefore, any solution of the equation
x + 3y = 4
will satisfy both equations simultaneously. So there are infinitely many solutions
because for any choice of value α for y, the equation x + 3y = 4 is satisfied by
x = 4− 3α, y = α.
In case (b), multiplying the first equation by 2 gives 2x+6y = 10, which is
inconsistent with the second equation. So the two given equations cannot have
any simultaneous solution.
For the case of three equations in three unknowns, a similar conclusion
holds. If the matrix of coefficients has zero determinant, then the equations
have either no simultaneous solutions or infinitely many. A method known as
Gauss-Jordan elimination can be applied to a system of equations to reduce it
to a simple form that eliminates redundant equations. The reduction is akin to
the usual process of simplifying equations; subtracting multiples of one equation
from another to eliminate variables between them. We shall say no more about
this in this book.
EXERCISES
10.1. Find the inverse, where it exists, of each of the matrices
5 2
6 3
,
3 −1
−15 5
,
3 −2
−4 3
,
−2 2
4 −3
,
3 0
0 5
,
0 1
1 0
.
232 Elements of Mathematics for Economics and Finance
10.2. The equilibrium prices P1, P2 for two goods satisfy the equations
4P1 − 3P2 = 11
−6P1 + 7P2 = 8.
Express these equations in matrix form. Hence, by inverting the
matrix, solve for P1, P2. Also solve these equations by Cramer’s
rule.
10.3. Let B =
3 2
0 −1
, C =
2 2
−1 −2
, D=
1 1
0 2
.
If B = ED, B = DF, and AB = AC + D, determine E, F, and A.
10.4. If XB = X +C, where B and C are as in the previous exercise, find
the 2 × 2 matrix X.
10.5. Determine which of the matrices
⎡
⎣
1 1 1
1 1 0
1 1 1
⎤
⎦ ,
⎡
⎣
1 1 1
1 1 0
1 0 0
⎤
⎦ ,
⎡
⎣
2 2 3
3 1 5
1 −7 3
⎤
⎦
are invertible.
(You do not have to find inverses.)
10.6. Find (i) |A|; (ii) the cofactor matrix of A; (iii) adj A, where A is the
matrix ⎡
⎣
2 5 3
4 5 2
7 7 1
⎤
⎦.
Hence find A−1.
10.7. Determine the equilibrium prices P1, P2, P3 of three interdependent
commodities that satisfy
2P1 + 5P2 + 3P3 = 136
4P1 + 5P2 + 2P3 = 132
7P1 + 7P2 + P3 = 160
using matrices or Cramer’s rule.
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