Application of Matrix and system

1. Business Mathematics Applications of Matrices and System of Linear Equation Dr. Litan Kumar Saha Professor Department of Applied Mathematics University of Dhaka - 1 -

2. 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.    a a a 11 12 1 n    a a a   21 22 2 n            a a a m 1 m 2 mn 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 - 3 -

3. 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   2000 2876 2314 II   7542 3214 2969 2. Number of staff in the office can be represented as follows:   2 Peon   4 Clerk     3 Typist   Head Clerk 1       Office Superintendent 1 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: Warehouse W2 Factory W1 W3 W4 I   13 12 17 14   II 22 26 11 19   III     16 15 18 11 - 4 -

4. 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 20 25 22 20   Cotton 10 20 18 10 A = Yarn       15 20 15 15 Year 2003 Jute   10 15 20 20   Cotton 5 20 18 10 B = Yarn       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     10 15 20 20 20 25 22 20     5 20 18 10 10 20 18 10 A + B = +             15 20 15 15 8 30 15 10   30 40 42 40   15 40 33 40 =       23 50 30 25 - 5 -

5. 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 B   50 45 A   Y = B 60 70 If the sales for the first three months is given as P Q B   30 15 A   Q = B 20 20 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 –     50 45 30 15     Y – Q = - 60 70 20 20 P Q B   20 30 A   = B 40 50 - 6 -

6. 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 Q R Ram   10  7 3  Shyam 4 8 10 Q = Mohan       4 7 8 × 3 3   P 4   5 Q C =       6 R × 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     4 8 10 5 Q × C = ×             4 7 8 6 - 7 -

7. × + × + ×   10 4 7 5 3 6   × + × + × 4 4 8 5 10 6 =       × + × + × 4 4 7 5 3 6 + +     40 35 18 93     + + 16 40 16 116 = =             + + 16 35 48 99 Amount spent by Ram, Shyam and Mohan is Rs 99, Rs 116 and Rs 99 respectively. - 8 -

8. 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   2 3 1 A   4 2 5 M = B C       2 4 2 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 A   2 3 1 A B C P Q R   [ ] = [ ] B 4 2 5 100 100 100 800 900 800       C 2 4 2 (ii). Let the per unit cost of materials P, Q and R be represented by the 3×1 matrix as under   10 Q R P  5  C =       5 With the help of matrix multiplication, the per unit cost of production of each product would be calculated as under   5   2 3 1     10 4 2 5 AC =             2 4 2 5 - 9 -

9. A   45   B 65 =   C     60 (iii). The total cost of production if the firm produces 200 units of each product would be given as   45   [ ] = [ ] 65 200 200 200 34 000 ,       60 Hence, the total cost of production will be Rs. 34,000. - 10 -

10. 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 = [ ] 1 3 10 Quantity Matrix = Q Shop Local Market   20 15 Sugar   Wheat 10 8 Price Matrix = P =   Salt     8 6   20 15   Therefore, Total Price = Q × P=[ ] 1 = [ ] 10 8 3 10 168 131       8 6 Now, Cost of purchasing from shop Cost of purchasing from local market= Rs 131 + Rs 25 (Cost of travel) = Rs 156 = Rs 168 and Hence, net savings to Mr. X from purchasing through Local Market = 168 – 156 = Rs 12 - 11 -

11. 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 + = y + = . 0 10 x . 0 15 y , 4 = 000 or 2 x 3 y 80 = 000 , + + . 0 12 x . 0 165 , 4 500 or 8 x 11 y , 3 00 000 , Expressing the above equations in matrix form, we obtain       2 3 x 80 000 , =  A X     , 3 B  8 11 y 00 000 , This can be written in the form AX = B or X = A-1B Since | A | = -2 ≠ 0, A-1exists and the solution can be given by: X = A-1B  − 2  , 20 − −      11 3 80 000 , 20 000 , 1 1 = − =  000 ,   , 3    − − 8 2 00 000 , 40 000 , 2  10 =   000 Hence x = Rs 10,000, y = Rs 20,000, and new investments would be Rs 12,000 and Rs 22,000 respectively. - 12 -

12. 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 c3 s1 2 3 4 Steel 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 x 3 + y 4 z 29 + x = y 2 z 13 = + + 3 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 ↔ R2the given system is equivalent to       1 1 2 x 13       = 2 3 4 y 29                   3 2 1 z 16 Now applying R2 → R2 – 2R1and R3 → R3 – 3R1, the above system is equivalent to       1 1 2 x 13       = 0 1 0 y 3                   − − − 0 1 5 z 23 Applying R3 → R3 + R2the above system is equivalent to - 13 -

13.       1 1 2 x 13       = 0 1 0 y 3                   − − 0 0 5 z 20 + + = ) 1 ( − x y 2 z 13 = ) 2 ( − y 3 − − = ) 3 ( − 5 z 20 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 -

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 + + = ) 1 ( − x y z 45 x = + z 8 − + + = ) 2 ( − . . e i x 0 y z 8 + = x z 2 y − + = ) 3 ( − . . e i x 2 y z 0 Therefore, we have, using (1), (2) and (3)       1 1 1 x 45       − 1 = 1 0 1 y 8                   − 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 ∆ = = 8 0 1 66 1 − 0 2 1 1 45 1 ∆ = − = 1 8 1 90 2 1 0 1 1 1 45 ∆ = − = 1 0 8 114 3 − 1 2 0 Therefore, - 15 -

15. 66 = = x 11 6 90 = = y 15 6 114 = = z 19 6 Hence, the production levels of the products are as follows: First product Second product Third product - 11 tons - 15 tons - 19 tons - 16 -