Paper XI Business Mathematics
PROJECT ON
Applications of Matrices to Business and Economics
Made By: Akash Gupta
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Contents Topic
Page No.
• Acknowledgements
2
• Introduction
4
• Application of Matrix Addition and Subtraction
6
• Application of Matrix Multiplication •
Application of System of Linear Equations
• Leontief’s Input-Output Model
8 13 18 25
• 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. a11 a 21 am1
a12 a22 am 2
a1n a2 n amn
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
I
Item B
2000 7542
II
Item C
2876 3214
2314 2969
2. Number of staff in the office can be represented as follows:
2 4 3 1 1
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: Factory I II III
W1
Warehouse W2 W3
W4
13 12 17 14 22 26 11 19 16 15 18 11
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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. Year 2002 QI Q2 Q3 Q4 Jute A = Cotton Yarn
20 25 22 20 10 20 18 10 15 20 15 15 Year 2003
Jute Cotton B= Yarn
10 15 20 20 5 20 18 10 8 30 15 10
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
=
20 25 22 20 10 20 18 10 15 20 15 15
=
30 40 42 40 15 40 33 40 23 50 30 25
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+
10 15 20 20 5 20 18 10 8 30 15 10
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 P Q
B50 45 60 70
A Y=
B
If the sales for the first three months is given as P
B30 15 20 20
A Q=
Q
B
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
=
50 45 30 15 60 70 - 20 20 P A
= B
Q
B 30 20 40 50
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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. P Ram Q = Shyam Mohan P C= Q R
Q
R
10 7 3 4 8 10 4 7 8 3× 3
4 5 6 3×1
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.
10 7 3 4 Q × C = 4 8 10 × 5 4 7 8 6 -7-
10 × 4 + 7 × 5 + 3× 6 = 4 × 4 + 8× 5 + 10 × 6 4 × 4 + 7 × 5 + 3× 6 40 + 35 + 18 = 16 + 40 + 16 16 + 35 + 48
93 = 116 99
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. P Q R A 2 3 1
4 2 5 2 4 2
M =B C
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. P
Q
R
2 3 1 A P Q R 4 2 5 B [100 100 100] = [ 800 900 800] 2 4 2C
A
(ii).
B
C
Let the per unit cost of materials P, Q and R be represented by the 3×1 matrix as under P 5 C
=Q R
10 5
With the help of matrix multiplication, the per unit cost of production of each product would be calculated as under
AC
2 3 1 5 = 4 2 5 10 2 4 2 5
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A =B C (iii).
45 65 60
The total cost of production if the firm produces 200 units of each product would be given as
45 [ 200 200 200] 65 = [ 34,000] 60 Hence, the total cost of production will be Rs. 34,000.
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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, Sugar Wheat Salt
Quantity Matrix
=Q
[
]
= 3 10 1 Shop Sugar
Price Matrix
=P
=Wheat Salt
Local Market
20 15 10 8 8 6
20 15 Therefore, Total Price = Q × P= [ 3 10 1] 10 8 8 6
= [168 131]
Now, Cost of purchasing from shop = Rs 168 and Cost 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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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 0.10 x + 0.15 y = 4,000 0.12 x + 0.165 y = 4,500
or or
2 x + 3 y = 80,000 8 x + 11 y = 3,00,000
Expressing the above equations in matrix form, we obtain 2 3 x 80,000 8 11 y = 3,00,000 A X B This 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 1 11 − 3 80,000 1 − 20,000 − =− 2 − 8 2 3,00,000 2 − 40,000 10,000 20,000
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: Cars c2
c1
Steel
c3
s1
2
3
4
s2
1
1
2
s3
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 2 x + 3 y + 4 z = 29 x + y + 2 z = 13 3 x + 2 y + z = 16 The above information can be represented using the matrix method, as under. 2 3 4 x 29 1 1 2 y = 13 3 2 1 z 16 The above equation can be solved using Gauss Jordan elimination method. By applying the operation R1 ↔ R2 the given system is equivalent to 1 1 2 x 13 2 3 4 y = 29 3 2 1 z 16 Now applying R2 → R2 – 2R1 and R3 → R3 – 3R1, the above system is equivalent to 2 x 13 1 1 0 1 0 y = 3 0 − 1 − 5 z − 23 Applying R3 → R3 + R2 the above system is equivalent to - 13 -
1 1 2 x 13 0 1 0 y = 3 0 0 − 5 z − 20 x + y + 2 z = 13 y=3 − 5 z = −20
−(1) −(2) −(3)
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 - 14 -
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 x + y + z = 45 z = x+8 i.e. − x + 0 y + z = 8 x + z = 2y i.e. x − 2y + z = 0
−(1) −(2) −(3)
Therefore, we have, using (1), (2) and (3) 1 1 x 45 1 − 1 0 1 y = 8 1 − 2 1 z 0 Which gives us 1 1 1 ∆ = −1 0 1 = 6 1 −2 1 Since ∆ ≠ 0, there is a unique solution. 45 1 1 ∆1 = 8 0 1 = 66 0 −2 1 1 45 1 ∆ 2 = − 1 8 1 = 90 1 0 1 1 1 45 ∆3 = −1 0 8 = 114 1 −2 0 Therefore,
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66 = 11 6 90 y= = 15 6 114 z= = 19 6 Hence, the production levels of the products are as follows: First product - 11 tons Second product - 15 tons Third product - 19 tons x=
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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: m11 p1 + m12 p2 + + m1n pn = p1 m21 p1 + m22 p 2 + + m2 n p n = p2 mn1 p1 + mn 2 p2 + + mnn p n = pn If m11 m A = 21 mn1
m12 m1n m22 m2 n mn 2 mnn
then the above system can be written as AP = P, where
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p1 p P = 2 pn 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: 0.3 p1 + 0.3 p 2 + 0.3 p3 = p1 0.4 p1 + 0.1 p 2 + 0.5 p3 = p 2 0.3 p1 + 0.6 p2 + 0.2 p3 = p3 The input-output matrix is 0.3 0.3 0.3 A = 0.4 0.1 0.5 0.3 0.6 0.2 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 0.3 0 − 0.7 0.3 0.4 − 0.9 0.5 0 0.3 0.6 − 0.8 0 which can be reduced to 1 0 − 0.82 0 0 1 − 0.92 0 0 0 0 0 To solve the system, we let p3 = t (a parameter), then the general solution is p1 = 0.82t p 2 = 0.92t p3 = t 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 p1 = 82 units p 2 = 92 units p3 = 100 units
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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 pi = mi1 p1 + mi 2 p 2 + + min p n + d i for each i. This gives the following linear system (written in a matrix form): P = AP + d where P and A are as above and d1 d d = 2 d 3 is the demand vector. One way to solve this linear system is
⇒
P = AP + d ( I − A) P = d
⇒
P = ( I − A) −1 d
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 0.1 025 0.2 A = 0.3 0.4 0.5 0.1 0.15 0.1 and the demand vector is 50,000 d = 75,000 1,25,000 Now, using the equation P = ( I − A) −1 d where 0.9 − 0.25 − 0.2 I − A = − 0.3 0.6 − 0.5 − 0.1 − 0.15 0.9 Using the Gauss Jordan elimination technique, we find that 1.464 0.803 0.771 ( I − A) −1 = 1.007 2.488 1.606 0.330 0.503 1.464
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Which gives 1.464 0.803 0.771 50,000 229921.59 P = 1.007 2.488 1.606 75,000 = 437795.27 0.330 0.503 1.464 1,25,000 237401.57
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 automanufacturing 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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