Jacques I. Mathematics for Economics and Business

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The firm charges both customers the same price for each product according to

P1 P2 P3

B = [100 500 200]

To make each item of type P1, P2 and P3, the firm uses four raw materials, R1, R2, R3 and R4. The

number of tonnes required per item is given by

R1 R2 R3 R4

C = P1

The cost per tonne of raw materials is

R1 R2 R3 R4

D = [20 10 15 15]

In addition, let

E = [1 1]

Find the following matrix products and give an interpretation of each one.

(a)

AB (b) AC (c) CD (d) ACD (e) EAB

(f) EACD (g) EAB − EACD

12 (1) Let

A = and B =

Find

(a)

(b) B

Do you notice any connection between (A + B)

, A

and B

(2) Let

C = and D =

Find

(a)

(b) D

Do you notice any connection between (CD)

, C

and D

13 Verify the equations

(a)

A(B + C) = AB + AC (b) (AB)C

A(BC)

in the case when

A = , B = and C =

−11

5 −3

210

−101

1 −1

−34

1001

1121

0011

Matrices

470

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14 If

A = [1 2 −43]andB =

find AB and BA.

15 Let

A = , A

−1

= (ad − bc ≠ 0)

I = and x =

Show that

(a)

AI = A and IA = A (b) A

−1

A = I and AA

−1

= I (c) Ix = x

(a) Evaluate the matrix product, Ax, where

A = and x =

Hence show that the system of linear equations

7x + 5y = 3

x + 3y = 2

can be written as Ax = b where b = .

(b) The system of equations

2x + 3y − 2z = 6

x − y + 2z = 3

4x + 2y + 5z = 1

can be expressed in the form Ax = b. Write down the matrices A, x and b.

d −b

−ca

ad − bc

−2

7.1 • Basic matrix operations

471

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section 7.2

Matrix inversion

In this and the following section we consider square matrices, in which the number of rows and

columns are equal. For simplicity we concentrate on 2 × 2 and 3 × 3 matrices, although the

ideas and techniques apply more generally to n × n matrices of any size. We have already seen

that, with one notable exception, the algebra of matrices is virtually the same as the algebra

of numbers. There are, however, two important properties of numbers which we have yet to

consider. The ﬁrst is the existence of a number, 1, which satisﬁes

a1 = a and 1a = a

for any number, a. The second is the fact that corresponding to any non-zero number, a, we

can ﬁnd another number, a

−1

, with the property that

−1

a = 1 and aa

−1

= 1

If you have worked through Practice Problem 15 of Section 7.1 you will know how to extend

these to 2 × 2 matrices. In part (a) you showed that, for any 2 × 2 matrix, A,

AI = A and IA = A

Objectives

At the end of this section you should be able to:

 Write down the 2 × 2 and 3 × 3 identity matrices.

 Detect whether a matrix is singular or non-singular.

 Calculate the determinant and inverse of a 2 × 2 matrix.

 Calculate the cofactors of a 3 × 3 matrix.

 Use cofactors to ﬁnd the determinant and inverse of a 3 × 3 matrix.

 Use matrix inverses to solve systems of linear equations arising in economics.

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where

I =

The matrix I is called the identity matrix and is analogous to the number 1 in ordinary arith-

metic. You also showed in part (b) of Practice Problem 15 that corresponding to the 2 × 2 matrix

A =

there is another matrix

−1

with the property that

−1

A = I and AA

−1

= I

The matrix A

−1

is said to be the inverse of A and is analogous to the reciprocal of a number.

The formula for A

−1

looks rather complicated but the construction of A

−1

is in fact very easy.

Starting with some matrix

A =

we ﬁrst swap the two numbers on the leading diagonal (that is, the elements along the line join-

ing the top left-hand corner to the bottom right-hand corner of A) to get

Secondly, we change the sign of the ‘off-diagonal’ elements to get

Finally, we multiply the matrix by the scalar

to get

The number ad − bc is called the determinant of A and is written as

det(A)or|A| or

Notice that the last step in the calculation is impossible if

|A |=0

because we cannot divide by zero. We deduce that the inverse of a matrix exists only if the

matrix has a non-zero determinant. This is comparable to the situation in arithmetic where a

reciprocal of a number exists provided the number is non-zero. If the matrix has a non-zero

determinant, it is said to be non-singular; otherwise it is said to be singular.

d −b

−

ad − bc

d −b

−

d −b

−

ad − bc

7.2 • Matrix inversion

473

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Matrices

474

Example

Find the inverse of the following matrices. Are these matrices singular or non-singular?

A = and B =

Solution

We begin by calculating the determinant of

A =

to see whether or not the inverse exists.

det(A) ==1(4) − 2(3) = 4 − 6 =−2

We see that det(A) ≠ 0, so the matrix is non-singular and the inverse exists. To ﬁnd A

−1

we swap the diag-

onal elements, 1 and 4, change the sign of the off-diagonal elements, 2 and 3, and divide by the determinant,

−2. Hence

−1

=− =

Of course, if A

−1

really is the inverse of A, then A

−1

A and AA

−1

should multiply out to give I. As a check:

−1

A ==✓

−1

==✓

To discover whether or not the matrix

B =

has an inverse we need to ﬁnd its determinant.

det(B) ==2(10) − 5(4) = 20 − 20 = 0

We see that det(B) = 0, so this matrix is singular and the inverse does not exist.

410

−21

3/2 −1/2

−21

3/2 −1/2

−21

3/2 −1/2

4 −2

−31

410

Practice Problem

1 Find (where possible) the inverse of the following matrices. Are these matrices singular or non-

singular?

A = B =

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7.2 • Matrix inversion

475

One reason for calculating the inverse of a matrix is that it helps us to solve matrix equations

in the same way that the reciprocal of a number is used to solve algebraic equations. We have

already seen in Section 7.1 how to express a system of linear equations in matrix form. Any

2 × 2 system

ax + by = e

cx + dy = f

can be written as

Ax = b

where

A = x = b =

The coefﬁcient matrix, A, and right-hand-side vector, b, are assumed to be given and the prob-

lem is to determine the vector of unknowns, x. Multiplying both sides of

Ax = b

by A

−1

gives

−1

(Ax) = A

−1

A)x = A

−1

b (associative property)

Ix = A

−1

b (deﬁnition of an inverse)

x = A

−1

b (Practice Problem 15(c) in Section 7.1)

The solution vector x can therefore be found simply by multiplying A

−1

by b. We are assuming

here that A

−1

exists. If the coefﬁcient matrix is singular then the inverse cannot be found and

the system of linear equations does not possess a unique solution; there are either inﬁnitely

many solutions or no solution.

Advice

These special cases are dealt with using the elimination method described in Section 1.2.

You might find it instructive to revise both Sections 1.2 and 1.3.

The following two examples illustrate the use of inverses to solve systems of linear equations.

The ﬁrst is taken from microeconomics and the second from macroeconomics.

Example

The equilibrium prices P

and P

for two goods satisfy the equations

−4P

+ P

=−13

− 5P

=−7

Express this system in matrix form and hence ﬁnd the values of P

and P



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476

Solution

Using the notation of matrices, the simultaneous equations

−4P

+ P

=−13

− 5P

=−7

can be written as

that is, as

Ax = b

where

A = x = b =

The matrix A has determinant

= (−4)(−5) − (1)(2) = 20 − 2 = 18

To ﬁnd A

−1

we swap the diagonal elements, −4 and −5, change the sign of the off-diagonal elements, 1 and

2, and divide by the determinant, 18, to get

−1

Finally, to calculate x we multiply A

−1

by b to get

x = A

−1

Hence P

= 4 and P

= 3.

−13

−7

−5 −1

−2 −4

−5 −1

−2 −4

−41

2 −5

−13

−7

−41

2 −5

−13

−7

−41

2 −5

Practice Problem

2 The equilibrium prices P

and P

for two goods satisfy the equations

+ P

= 43

+ 7P

= 57

Express this system in matrix form and hence find the values of P

and P

[You have already solved this particular system in Practice Problem 4 of Section 1.3. You might like

to compare the work involved in solving this system using the method of elimination described in

Chapter 1 and the method based on matrix inverses considered here.]

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7.2 • Matrix inversion

477

Example

The equilibrium levels of consumption, C, and income, Y, for the simple two-sector macro-economic model

satisfy the structural equations

Y = C + I*

C = aY + b

where a and b are parameters in the range 0 < a < 1 and b > 0, and I* denotes investment. Express this sys-

tem in matrix form and hence express Y and C in terms of a, b and I*. Give an economic interpretation of

the inverse matrix.

Solution

The reduced form of the structural equations for this simple model has already been found in Section 5.3.

It is instructive to reconsider this problem using matrices. The objective is to express the endogenous vari-

ables, Y and C, in terms of the exogenous variable I* and parameters a and b. The ‘unknowns’ of this prob-

lem are therefore Y and C, and we begin by rearranging the structural equations so that these variables

appear on the left-hand sides. Subtracting C from both sides of

Y = C + I*

gives

Y − C = I* (1)

and if we subtract aY from both sides of

C = aY + b

we get

−aY + C = b (2)

(It is convenient to put the term involving Y ﬁrst so that the variables align with those of equation (1).)

In matrix form, equations (1) and (2) become

that is,

Ax = b

where

A = x = b =

The matrix A has determinant

= 1(1) − (−1)(−a) = 1 − a

which is non-zero because a < 1.

To ﬁnd A

−1

, we swap the diagonal elements, 1 and 1, change the sign of the off-diagonal elements, −1 and

−a, and divide by the determinant, 1 − a, to get

−1

a 1

1 − a

1 −1

−a 1

1 −1

−a 1

1 −1

−a 1



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478

Finally, to determine x we multiply A

−1

by b to get

x = A

−1

Hence

Y = and C =

The inverse matrix obviously provides a useful way of solving the structural equations of a macroeconomic

model. In addition, the elements of the inverse matrix can be given an important economic interpretation.

To see this, let us suppose that the investment I* changes by an amount ∆I* to become I* +∆I*, with the

parameter b held ﬁxed. The new values of Y and C are obtained by replacing I* by I* +∆I* in the expres-

sions for Y and C, and are given by

and

respectively. The change in the value of Y is therefore

∆Y =−=∆I*

and the change in the value of C is

∆C =−=∆I*

In other words, the changes to Y and C are found by multiplying the change in I* by

and

respectively. For this reason we call

the investment multiplier for Y and

the investment multiplier for C.

Now the inverse matrix is

−1

and we see that these multipliers are precisely the elements that appear in the ﬁrst column. It is easy to show,

using a similar argument, that the second column contains the multipliers for Y and C due to changes in the

autonomous consumption, b. The four elements in the inverse matrix can thus be interpreted as follows:

I* b

investment multiplier for Y autonomous consumption multiplier for Y

investment multiplier for C autonomous consumption multiplier for C

1 − a 1 − a

a 1

1 − a 1 − a

1 − a

aI* + b

1 − a

a(I* +∆I *) + b

1 − a

I* + b

1 − a

I* +∆I * + b

1 − a

a(I* +∆I *) + b

1 − a

I* +∆I* + b

1 − a

aI* + b

1 − a

I* + b

1 − a

I* + b

aI* + b

1 − a

a 1

1 − a

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7.2 • Matrix inversion

479

The concepts of determinant, inverse and identity matrices apply equally well to 3 × 3 matrices.

The identity matrix is easily dealt with. It can be shown that the 3 × 3 identity matrix is

I =

You are invited to check that, for any 3 × 3 matrix A,

AI = A and IA = A

Before we can discuss the determinant and inverse of a 3 × 3 matrix we need to introduce an

additional concept known as a cofactor. Corresponding to each element a

of a matrix A, there

is a cofactor, A

. A 3 × 3 matrix has nine elements, so there are nine cofactors to be computed.

The cofactor, A

, is deﬁned to be the determinant of the 2 × 2 matrix obtained by deleting row

i and column j of A, preﬁxed by a ‘+’ or ‘−’ sign according to the following pattern

For example, suppose we wish to calculate A

, which is the cofactor associated with a

in the matrix

A =

The element a

lies in the second row and third column. Consequently, we delete the second

row and third column to produce the 2 × 2 matrix

The cofactor, A

, is the determinant of this 2 × 2 matrix preﬁxed by a ‘−’ sign because from the

pattern

+−+

−+−

+−+

−+−

+−+

100

010

001

Practice Problem

3 The general linear supply and demand equations for a one-commodity market model are given by

P = aQ

+ b (a > 0, b > 0)

P =−cQ

+ d (c > 0, d > 0)

Show that in matrix notation the equilibrium price, P, and quantity, Q, satisfy

Solve this system to express P and Q in terms of a, b, c and d. Write down the multiplier for Q due to

changes in b and deduce that an increase in b leads to a decrease in Q.

1 −a

1 c

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