Paper XIBusiness Mathematics

PROJECTON

Applications of Matrices to Business and Economics

Made By: Akash Gupta

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Contents

Topic Page No.

• Acknowledgements

• Introduction

• Application of Matrix Addition and Subtraction

• Application of Matrix Multiplication

• Application of System of Linear Equations

• Leontief’s Input-Output Model

• Bibliography

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Applications of Matrices to

Business and Economics

What is a Matrix?

A matrix is a two-dimensional arrangement of numbers in rows and columns enclosed by a pair of square brackets ([ ]), in the form shown below.

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

The above figure shows an m × n matrix of m rows and n columns.

Matrices are used to describe linear equations, keep track of the coefficients of linear transformations and to record data that depend on multiple parameters. They can be added, multiplied, and decomposed in various ways, which also makes them a key concept in the field of linear algebra.

The subject of matrices has been researched and expanded by the works of many mathematicians, who have found numerous applications of matrices in various disciplines such as Economics, Engineering, Statistics and various other sciences.

In this project, the following applications to matrices will be discussed:

• Applications of Matrix Addition and Subtraction• Applications of Multiplication of Matrices• Applications of System of Linear Equations• Leontief Input-Output Model

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But first, let’s discuss how various situations in business and economics can be represented using matrices. This can be done using the following examples.

1. Annual productions of two branches selling three types of items may be represented as follows:

Branch Item A Item B Item C

I

II

2. Number of staff in the office can be represented as follows:

Peon

Clerk

Typist

Head Clerk

Office Superintendent

3. The unit cost of transportation of an item from each of the three factories to each of the four warehouses can be represented as follows:

WarehouseFactory W1 W2 W3 W4

I

II

III

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296932147542

231428762000

1

1

3

4

2

11181516

19112622

14171213

Applications of Matrix Addition and Subtraction

The applications of addition and subtraction of matrices can be illustrated through the following examples:

Illustration 1 - The quarterly sales of Jute, Cotton and Yarn for the year 2002 and 2003 are given below.

A =

15152015

10182010

20222520

B =

1015308

1018205

20201510

Find the total quarterly sales of Jute, Cotton and Yarn for the two years.

Solution –

The total sales of Jute, Cotton and Yarn will be obtained as under

A + B =

15152015

10182010

20222520

+

1015308

1018205

20201510

=

25305023

40334015

40424030

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Jute

Cotton

Yarn

Jute

Cotton

Yarn

QI Q2 Q3 Q4

Year 2002

Year 2003

Illustration 2 – X Ltd has the following sales position of its products A and B at its two centers P and Q at the end of the year

Y =

7060

4550

If the sales for the first three months is given as

Q =

2020

1530

Find the sales position for the last nine months.

Solution –

Given are the sales positions for the whole year (Y) and for the first three months (Q).

Hence, sales position for the remaining nine months –

Y – Q =

7060

4550 -

2020

1530

=

5040

3020

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A

B

P Q

B

A

B

P Q

B

A

B

P Q

B

Applications of Matrix Multiplication

It is important to note that two matrices can be multiplied if and only if the number of columns of the first matrix equals the number of rows of the second. The resultant matrix will have the number of rows equal to the first matrix and number of columns equal to that of the second matrix.

In other words,• A matrix of the order [a x b] can only be multiplied with a matrix of order [b x c]• The resultant matrix will be of the order [a x c]

The application of multiplication of matrices can be illustrated through the following examples.

Illustration 3 – Ram, Shyam and Mohan purchased biscuits of different brands P, Q and R. Ram purchased 10 packets of P, 7 packets of Q and 3 packets of R. Shyam purchased 4 packets of P, 8 packets of Q and 10 packets of R. Mohan purchased 4 packets of P, 7 packets of Q and 8 packets of R. If brand P costs Rs 4, Q costs Rs 5 and R costs Rs 6 each, then using matrix operation, find the amount of money spent by these persons individually.

Solution – Let Q be the matrix denoting the quantity of each brand of biscuit bought by P, Q and R and let C be the matrix showing the cost of each brand of biscuit.

Q =

33874

1084

3710

×

C =

136

5

4

×

Since number of columns of first matrix should be equal to the number of rows of the second matrix for multiplication to be possible, the above matrices shall be multiplied in the following order.

Q × C =

874

1084

3710

×

6

5

4

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Ram

Shyam

Mohan

P Q R

P

Q

R

=

×+×+××+×+××+×+×

635744

6105844

6357410

=

++++++

483516

164016

183540

=

99

116

93

Amount spent by Ram, Shyam and Mohan is Rs 99, Rs 116 and Rs 99 respectively.

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Illustration 4- A firm produces three products A, B and C requiring the mix of three materials P, Q and R. The requirement (per unit) of each product for each material is as follows.

M =

242

524

132

Using matrix notations, find(i). The total requirement of each material if the firm produces 100 units of each

product.(ii). The per unit cost of production of each product if the per unit cost of materials

P, Q and R is Rs 5, Rs 10 and Rs 5 respectively.(iii). The total cost of production if the firm produces 200 units of each product.

Solution –

(i). The total requirement of each material if the firm produces 100 units of each product can be calculated using the matrix multiplication given below.

[ ]100100100

242

524

132

= [ ]800900800

(ii). Let the per unit cost of materials P, Q and R be represented by the 3×1 matrix as under

C =

5

10

5

With the help of matrix multiplication, the per unit cost of production of each product would be calculated as under

AC =

242

524

132

5

10

5

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A

B

C

P Q R

P Q R

A

B

C

A B C P Q R

P

Q

R

=

60

65

45

(iii). The total cost of production if the firm produces 200 units of each product would be given as

[ ]200200200

60

65

45

= [ ]000,34

Hence, the total cost of production will be Rs. 34,000.

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A

B

C

Illustration 5 - Mr. X went to a market to purchase 3 kg of sugar, 10 kg of wheat and 1 kg of salt. In a shop near to Mr. X’s residence, these commodities are priced at Rs 20, Rs 10 and Rs 8 per kg whereas in the local market these commodities are priced at Rs 15, Rs 8 and Rs 6 per kg respectively. If the cost of traveling to local market is Rs 25, find the net savings of Mr. X, using matrix multiplication method.

Solution – Let matrices Q and P represent quantity and price. Then,

Quantity Matrix = Q = [ ]1103

Price Matrix = P =

68

810

1520

Therefore, Total Price = Q × P= [ ]1103

68

810

1520

= [ ]131168

Now,Cost of purchasing from shop = Rs 168 andCost of purchasing from local market= Rs 131 + Rs 25 (Cost of travel) = Rs 156

Hence, net savings to Mr. X from purchasing through Local Market= 168 – 156 = Rs 12

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Sugar Wheat Salt

Shop Local Market

Sugar

Wheat

Salt

Applications of System of Linear Equations

The following examples can be used to illustrate the common methods of solving systems of linear equations that result from applied business and economic problems.

Illustration 6 – Mr. X invested a par of his investment in 10% bond A and a part in 15% bond B. His interest income during the first year is Rs 4,000. If he invests 20% more in 10% bond A and 10 % more in 15% bond B, his income during the second year increases by Rs 500. Find his initial investment and the new investment in bonds A and B using matrix method.

Solution – Let initial investment be x in 10% bond A and y in 15% bond B. Then, according to given information, we have

000,00,3118or500,4165.012.0

000,8032or000,415.010.0

=+=+=+=+

yxyx

yxyx

Expressing the above equations in matrix form, we obtain

=

000,00,3

000,80

118

32

y

x

A X BThis can be written in the form AX = B or X = A-1B

Since | A | = -2 ≠ 0, A-1 exists and the solution can be given by:

X = A-1B

=

−−

−=

−

−−

000,40

000,20

2

1

000,00,3

000,80

28

311

2

1

=

000,20

000,10

Hence x = Rs 10,000, y = Rs 20,000, and new investments would be Rs 12,000 and Rs 22,000 respectively.

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Illustration 7 – An automobile company uses three types of steel s1, s2 and s3 for producing three types of cars c1, c2 and c3. The steel requirement (in tons) for each type of car is given below:

Carsc1 c2 c3

s1

Steel s2

s3

2 3 4

1 1 2

3 2 1

Determine the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of the three types respectively.

Solution – Let x, y and z denote the number of cars that can be produced of each type. Then we have

1623

132

29432

=++=++

=++

zyx

zyx

zyx

The above information can be represented using the matrix method, as under.

=

16

13

29

123

211

432

z

y

x

The above equation can be solved using Gauss Jordan elimination method.By applying the operation R1 ↔ R2 the given system is equivalent to

=

16

29

13

123

432

211

z

y

x

Now applying R2 → R2 – 2R1 and R3 → R3 – 3R1, the above system is equivalent to

−=

−− 23

3

13

510

010

211

z

y

x

Applying R3 → R3 + R2 the above system is equivalent to

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−=

− 20

3

13

500

010

211

z

y

x

)3(205

)2(3

)1(132

−−=−−=−=++

z

y

zyx

Therefore, z = 4. Substituting y = 3 and z = 4 in (1), we get x = 2. Hence the solution is

x = 2, y = 3 and z = 4

Illustration 8 – A company produces three products everyday. Their total production on a certain day is 45 tons. It is found that the production of the third product exceeds the

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production of the first product by 8 tons while the total combined production of the first and third product is twice that of the second product. Determine the production level of each product using Cramer’s rule.

Solution – Let the production level of the three products be x, y and z respectively. Therefore, we will have the following equations

)3(02..

2

)2(80..

8

)1(45

−=+−=+

−=++−+=

−=++

zyxei

yzx

zyxei

xz

zyx

Therefore, we have, using (1), (2) and (3)

=

−−

0

8

45

121

101

111

z

y

x

Which gives us

6

121

101

111

=−

−=∆

Since ∆ ≠ 0, there is a unique solution.

66

120

108

1145

1 =−

=∆

90

101

181

1451

2 =−=∆

114

021

801

4511

3 =−

−=∆

Therefore,

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196

114

156

90

116

66

==

==

==

z

y

x

Hence, the production levels of the products are as follows:First product - 11 tonsSecond product - 15 tonsThird product - 19 tons

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The Leontief Input-Output Model

The Leontief Input-Output Model discusses the interdependence of industries on each other. Based on the assumption that each industry in the economy has two types of demands: external demand (from outside the system) and internal demand (demand placed on one industry by another in the same system), the Leontief model represents the economy as a system of linear equations. The Leontief model was invented in the 30’s by Professor Wassily Leontief who was awarded the Nobel Prize in Economics in 1973 for his effort.

There are two types of Leontief models, i.e. Closed and Open.

Closed Input-Output Model

Consider an economy consisting of n interdependent industries (or sectors) S1,…,Sn. That means that each industry consumes some of the goods produced by the other industries, including itself (for example, a power-generating plant uses some of its own power for production). We say that such an economy is closed if it satisfies its own needs; that is, no goods leave or enter the system. Let mij be the number of units produced by industry Si

and necessary to produce one unit of industry Sj. If pk is the production level of industry Sk, then mij pj represents the number of units produced by industry Si and consumed by industry Sj . Then the total number of units produced by industry Si is given by:

p1mi1+p2mi2+…+pnmin.

In order to have a balanced economy, the total production of each industry must be equal to its total consumption. This gives the linear system:

nnnnnn

nn

nn

ppmpmpm

ppmpmpm

ppmpmpm

=+++

=+++=+++

2211

22222121

11212111

If

=

nnnn

n

n

mmm

mmm

mmm

A

21

22221

11211

then the above system can be written as AP = P, where

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=

np

p

p

P

2

1

A is called the input-output matrix.

We are then looking for a vector P satisfying AP = P and with non-negative components, at least one of which is positive.

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Illustration 9 - Suppose that the economy of a certain region depends on three industries: service, electricity and oil production. Monitoring the operations of these three industries over a period of one year, we were able to come up with the following observations:

1. To produce 1 unit worth of service, the service industry must consume 0.3 units of its own production, 0.3 units of electricity and 0.3 units of oil to run its operations.

2. To produce 1 unit of electricity, the power-generating plant must buy 0.4 units of service, 0.1 units of its own production, and 0.5 units of oil.

3. Finally, the oil production company requires 0.3 units of service, 0.6 units of electricity and 0.2 units of its own production to produce 1 unit of oil.

Find the production level of each of these industries in order to satisfy the external and the internal demands assuming that the above model is closed, that is, no goods leave or enter the system.

Solution

Consider the following variables:

1. p1= production level for the service industry

2. p2= production level for the power-generating plant (electricity)

3. p3= production level for the oil production company

Since the model is closed, the total consumption of each industry must equal its total production. This gives the following linear system:

3321

2321

1321

2.06.03.0

5.01.04.0

3.03.03.0

pppp

pppp

pppp

=++=++=++

The input-output matrix is

=

2.06.03.0

5.01.04.0

3.03.03.0

A

and the above system can be written as (A-I)P = 0. Note that this homogeneous system has infinitely many solutions (and consequently a nontrivial solution) since each column

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in the coefficient matrix sums to 1. The augmented matrix of this homogeneous system is

−−

−

08.06.03.0

05.09.04.0

03.03.07.0

which can be reduced to

−−

0000

092.010

082.001

To solve the system, we let p3 = t (a parameter), then the general solution is

tp

tp

tp

===

3

2

1

92.0

82.0

and as we mentioned above, the values of the variables in this system must be nonnegative in order for the model to make sense. In other words, t ≥ 0. Taking t=100 for example would give the solution

unitsp

unitsp

unitsp

100

92

82

3

2

1

===

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Open Input-Output Model

The first Leontief model treats the case where no goods leave or enter the economy, but in reality this does not happen very often. Usually, a certain economy has to satisfy an outside demand, for example, from bodies like the government agencies. In this case, let di be the demand from the ith outside industry, pi, and mij be as in the closed model above, then

ininiii dpmpmpmp ++++= 2211

for each i. This gives the following linear system (written in a matrix form):

dAPP +=

where P and A are as above and

=

3

2

1

d

d

d

d

is the demand vector.

One way to solve this linear system is

dAIP

dPAI

dAPP

1)(

)(−−=⇒

=−⇒+=

Of course, we require here that the matrix I-A be invertible, which might not be always the case. If, in addition, (I-A)-1 has nonnegative entries, then the components of the vector P are nonnegative and therefore they are acceptable as solutions for this model. We say in this case that the matrix A is productive.

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Illustration 10 - Consider an open economy with three industries: coal-mining operation, electricity-generating plant and an auto-manufacturing plant. To produce Re 1 of coal, the mining operation must purchase Re 0.1 of its own production, Rs 0.30 of electricity and Re 0.1 worth of automobile for its transportation. To produce Re 1 of electricity, it takes Rs 0.25 of coal, Rs 0.4 of electricity and Rs 0.15 of automobile. Finally, to produce Re 1 worth of automobile, the auto-manufacturing plant must purchase Rs 0.2 of coal, Rs 0.5 of electricity and consume Rs 0.1 of automobile. Assume also that during a period of one week, the economy has an exterior demand of Rs 50,000 worth of coal, Rs 75,000 worth of electricity, and Rs 1,25,000 worth of autos. Find the production level of each of the three industries in that period of one week in order to exactly satisfy both the internal and the external demands.

Solution

The input-output matrix of this economy is

=

1.015.01.0

5.04.03.0

2.00251.0

A

and the demand vector is

=

000,25,1

000,75

000,50

d

Now, using the equation

dAIP 1)( −−=

where

−−−−−−

=−9.015.01.0

5.06.03.0

2.025.09.0

AI

Using the Gauss Jordan elimination technique, we find that

=− −

464.1503.0330.0

606.1488.2007.1

771.0803.0464.1

)( 1AI

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Which gives

=

=

57.237401

27.437795

59.229921

000,25,1

000,75

000,50

464.1503.0330.0

606.1488.2007.1

771.0803.0464.1

P

So, the total output of the coal-mining operation must be Rs 229921.59, the total output for the electricity-generating plant is Rs 437795.27 and the total output for the auto-manufacturing plant is Rs 237401.57.

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BIBLIOGRAPHY

1. ‘Business Mathematics’ by S. K. Singh and J. K. Singh, Brijwasi Book Distributors and Publishers

2. ‘Mathematics for Business Studies’ by Dr. J. K. Thukral, Mayur Paperbacks

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