Matrix Algebra and its Applications

UNIT 1 107 Basic mathematics for management 4 Matrix Algebra and its Applications Objectives By the end of this section, you should be able to: 1. Explain the basic concepts of a matrix. 2. Apply methods of representing large quantities of data in matrix form. 3. Perform various operations concerning matrices. 4. Find the solution methods of simultaneous linear equations. 5. Discuss applications of matrix algebra in various decision models. Introduction Matrices have proven their usefulness in quantitative analysis of managerial decisions in several disciplines like marketing, finance, production, personnel, economics, etc. Many quantitative methods such as linear programming, game theory, Markov models, input-output models and some statistical models have matrix algebra as their underlying theoretical base. All these models are built by establishing a system of linear equations which represent the problems to be solved. The simultaneous linear equations involving more than three variables cannot be solved by using “ordinary algebra”. Real-world business problems may involve more than three variables, and in such cases matrices are used to represent a complex system of equations and large quantities of data in a compact form. Once the system of equations is represented in matrix form, they can be solved easily and quickly by using a computer. The limitation of matrix algebra is that it is applicable only in those cases where assumption of linearity can be made. The main objective of this section is to provide (1) some basic theoretical matrix operations, namely addition, subtraction, and multiplication (2) a procedure for solving a system of linear simultaneous equations, and (3) a few applications of matrix algebra. 108 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques 4.1 Matrices: Definition and notations A matrix is a rectangular array of ordered numbers. The term ordered implies that the position of each number is significant and must be determined carefully to represent the information contained in the problem. These numbers (also called elements of the matrix) are arranged in rows and columns of the rectangular array and are enclosed by either square brackets [ ], parentheses ( ) or by a pair of double vertical lines || ||. A matrix consisting of m rows and n columns is written in the following form. a column a11 a21 . . am1 am2 … amn a12 a22 … … a1n a2n a row where a11, a12, … denote the numbers (or elements) of the matrix. The dimension (or order) of the matrix is determined by the number of rows and columns. Here, in the given matrix, there are m rows and n columns. Therefore, it is of the dimension m × n (read as m by n). In the dimension of the given matrix, the number of rows is always specified first and followed by the number of columns. Boldface capital letters such as A, B, C ... are used to denote the entire matrix. The matrix is also sometimes represented as A = [aij]mxn where aij denotes the element in the ith row and the jth element of A. Some examples of the matrices are A= [ ] −1 1 ; 2 3 2×2 B= [ 1 1 2 ; 2 4 −3 2×3 ] 5 5 10 C = 6 2 10 2 1 2 [ ] 3×3 The matrix A is a 2 × 2 matrix because it has 2 rows and 2 columns. Similarly, the matrix B is a 2 × 3 matrix while matrix C is a 3 × 3 matrix. UNIT 1 109 Basic mathematics for management Activity A Tick the correct alternative indicating the dimension of the matrix [ ] 2 6 3 3 8 5 4 9 7 A) 3 × 3 B) 4 × 3 3×3 C) None of these. 4.2 Some special matrices Square matrix A matrix in which the number of rows equals the number of columns is called a square matrix. For example [ ] 2 6 3 3 8 5 4 9 7 3×3 is a square matrix of dimension 3. The elements 2, 8 and 7 in this matrix are called the diagonal elements and the diagonal is called the principal diagonal. Diagonal matrix A square matrix, in which all non-diagonal elements are zero whereas diagonal elements are non-zero, is called a diagonal matrix. For example [ ] 2 0 0 0 5 0 0 0 1 3×3 is a diagonal matrix of dimension 3. 110 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Scalar matrix A diagonal matrix in which all diagonal elements are equal is called a scalar matrix. For example [ ] k 0 0 0 0 k 0 0 k 3×3 is a scalar matrix, where k is a real (or complex) number. Identity (or unit) matrix A scalar matrix in which all diagonal elements are equal to one, is called an identity (or unit) matrix and is denoted by I. Following are two different identity matrices I2 = [ ] 1 0 ; 0 1 2×2 1 0 0 I3 = 0 1 0 0 0 1 [ ] 3×3 An identity matrix of dimension n is denoted by In. It has n elements in its diagonal, each equal to 1 and other elements are zero. The zero (or null) matrix A matrix is said to be the zero matrix if every element of it is zero. It is denoted as 0. Following are three different zero matrices: [ ] 0 0 ; 0 0 2×2 [ 0 0 0 ; 0 0 0 2×3 ] [ ] 0 0 0 0 0 0 3×2 4.3 Matrix representation of data Before discussing the operations on matrices, it is necessary for you to know a few situations in which data can be represented in matrix form. UNIT 1 111 Basic mathematics for management Transportation problem The unit cost of transportation of an item from each of the two factories to each of the three warehouses can be represented in a matrix as shown below: Warehouses W1 Factory F1 F2 W2 15 20 W3 30 15 [ 20 25 ] Similarly, we can also construct a time matrix [tij], where tij = time of transportation of an item from factory i to warehouse j. Note that the time of transportation is independent of the amount shipped. Distance matrix The distance (in km) between the given number of cities can be represented as a matrix as shown below: City A A City B C D B 1,470 – 1,853 2,365 C 2,158 1,853 – 1,635 D 1,732 2,385 1,635 – [ – 1,470 2,158 1,732 ] Diet matrix The vitamin content of two types of food and two types of vitamins can be represented in a matrix as shown below: Vitamins A Food F1 F2 B 120 100 [ 150 170 ] 112 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Assignment matrix The time required to perform three jobs by three workers can be represented in a matrix as shown below: Job J1 W1 Worker W2 W3 J2 3 5 4 J3 2 3 6 [ 5 4 2 ] Pay-off matrix Suppose two players A and B play a coin tossing game. If the outcome of (H, H) or (T, T) occurs, then player B loses RM10 to player A, otherwise player B gains RM10 from player A as shown in the matrix: Player B H Player A H T T −10 10 [ 10 −10 ] The minus sign with the pay-off means that player A pays to player B. Brand switching matrix The proportion of users in the population surveyed switching from brand i to brand j of an item in a period can be represented as a matrix: To Brand 1 Brand 1 From Brand 2 Brand 3 Brand 2 0.6 0.3 0.5 Brand 3 0.1 0.1 0.3 [ 0.3 0.6 0.2 ] Here the sum of the elements of each row is 1 because these are the proportions. UNIT 1 113 Basic mathematics for management 4.4 Operations on matrices Addition (or subtraction) of matrices The addition (or subtraction) of two or more matrices is possible only if these matrices have the same dimensions, i.e., matrices must have the same number of rows and the same number of columns. The sum (or difference) of matrices is obtained by adding (or subtracting) the corresponding elements of the given matrices. For example, if A= [ ] 1 3 2 4 and B= 2×2 [ ] −1 7 0 8 0 2 2×2 then A+B= [ [ 1−1 3+7 2+0 4+8 ] = 2×2 [ ] 10 12 = 2 2 2×2 A−B= 1 − (−1) 2−0 3−7 4−8 ] [ ] −4 −4 Note that A − B ≠ B − A. Example 1 A company produces three types of products A, B and C. The total annual sales (in ‘000 units) of these products for the years 2005 and 2006 in the four regions are given below. For the year 2005: Region Product A B C Eastern 15 5 8 Western 8 24 4 Southern 6 7 31 Northern 12 8 6 114 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques For the year 2006: Region Product A B C Eastern 17 5 13 Western 10 22 6 Southern 5 11 39 Northern 7 4 5 Find the total sales of three products for two years. Solution: The total sales of three products for two years can be obtained by adding the sales of two years as shown below: Region Product A B C Eastern 15 + 17 = 32 5 + 5 = 10 8 + 13 = 21 Western 8 + 10 = 18 24 + 22 = 46 4 + 6 = 10 Southern 5 + 5 = 10 7 + 11 = 18 31 + 39 = 70 Northern 12 + 7 = 19 8 + 4 = 12 5 + 6 = 11 Properties of matrix addition If A, B and C are any three matrices of same dimension, then 1. Matrix addition is commutative, i.e., A + B = B + A. 2. Matrix addition is associative, i.e., (A + B) + C = A + (B + C). 3. For any matrix A of dimension m × n, there is a zero matrix of the same dimension such that A + 0 = 0 + A = A. This shows that zero matrix is the additive identity. UNIT 1 115 Basic mathematics for management 4. If for any matrix A of dimension m × n, there exists another matrix B of the same dimension such that A+B=B+A=O then B is called the additive inverse (or negative) of A and is denoted by −A. Activity B If matrices A and B are defined as A= [ 0 2 3 ; 2 1 4 ] B= [ 7 1 6 4 3 5 ] then compute: a) A + B b) A − B c) B − A Scalar multiplication If A= [aij] is any matrix of dimension m × n and k is any scalar (real number), then the multiplication kA is obtained by simply multiplying each element of A by the scalar k. That is Ak = kA = [kaij] Example 2 The sales figures in Example 1 are given in thousands of units. If we want to express the sales figures in actual units, then we have to multiply the given matrices by 1000. For illustration, let us consider the data matrix of 1985. That is, if A= [ 15 5 8 8 24 4 5 7 31 12 8 6 ] 116 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques then 15 × 1000 1000 A = 5 × 1000 8 × 1000 15,000 = 5,000 8,000 [ [ 8 × 1000 24 × 1000 4 × 1000 8,000 24,000 4,000 5 × 1000 7 × 1000 31 × 1000 12,000 8,000 6,000 12 × 1000 8 × 1000 6 × 1000 ] 5,000 7,000 31,000 ] Properties of scalar multiplication 1. k(A + B) = kA + kB where A and B are two matrices of same dimension and k is a scalar number. 2. (k1 + k2) A = k1A + k2A where A is a matrix and k1 and k2 are two distinct scalar numbers. Activity C If two matrices A and B are defined as A= [ 0 2 3 ; 2 1 4 ] B= [ 7 1 6 4 3 5 ] then compute 2A + 3B. Multiplication of matrices The matrix multiplication consists of the following steps: 1. Check on compatibility: The following dimensional arrangement must hold for compatibility in matrix multiplication: Dimensions: lead matrix × lag matrix = product (m × p) × (p × n) = m × n In other words, the number of columns in the first matrix must be equal to the number of rows in the second matrix. If this condition does not exist, then the matrices are said to be incompatible and their multiplication is not defined. UNIT 1 117 Basic mathematics for management 2. The operation of multiplication: For multiplication of two matrices, the following procedure should be adopted: a) The element of a row of the lead matrix A should be multiplied by the corresponding elements of a column of the lag matrix B. b) The product is then added and the location of this resulting element in the new matrix C determines which row from A has to be multiplied with which column from B. To illustrate this, let us take two matrices A and B as defined below: A= [ × 2 3 3 5 ; 5 7 2×3 ] 2 5 B= 3 5 5 7 [ ] 3×2 then A (2 × 3) R1 R2 B (3 × 2) 2 3 5 5 5 7 = C (2 × 2) [ 2 3 3 5 5 7 ] × [ ][ C1 C2 = (2)(2) + (3)(3) + (5)(5) (2)(5) + (3)(5) + (5)(7) (3)(2) + (5)(3) + (7)(5) (3)(5) + (5)(5) + (7)(7) ] = 38 56 [ 60 89 ] 2×2 Example 3 There are two families A and B. There are 2 men, 3 women and 1 child in family A and 1 man, 1 woman and 2 children in family B. The recommended daily allowance for calories is: man, 2400; woman, 1900; child, 1800, and for proteins: man, 55 gm; woman, 45 gm and child, 33 gm. Represent the above information by matrices. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. Solution: man Let C = Family A B 2 1 women 3 1 child 1 2 2×3 and 118 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Calory D = Man Woman Child Protein 55 45 33 [ 2400 1900 1800 ] 3×2 If you look at the dimensions of two matrices C and D you will find that the condition for multiplication is satisfied. Therefore, the total requirement of calories and proteins for each of the two families is determined by multiplying C and D, as shown below: R1 R2 C×D [ 2400 55 2 3 1 R 1C 1 R 1C 2 × 1900 45 = 1 1 2 R2C1 R2C2 1800 33 ] [ ][ ] 2×2 C1 Calory Family A B C2 Protein 278 166 [ 12300 7900 ] 2×2 Activity D 1. If two matrices of dimension m × n and n × p are multiplied, then the resulting matrix is of dimension: A) m × n B) n × p C) m × p D) None of these. 2. If A and B are two non-zero compatible matrices with respect to multiplication, then their product A) is always zero matrix. B) is never a zero matrix. C) may be a zero matrix. D) None of these. UNIT 1 119 Basic mathematics for management 3. A factory employs 50 skilled workers and 20 unskilled workers. The daily wages to skilled and unskilled workers are RM30 and RM17 respectively. Using matrix notation, find: a) The matrix representing the number of workers. b) The total daily payment made to the workers. Properties of matrix multiplication 1. Matrix multiplication, in general, is not commutative, i.e., AB ≠ BA. 2. Matrix multiplication is associative, i.e., A(BC) = (AB)C where A, B, C are any three matrices of dimension m × n, n × p and p × q respectively. 3. Matrix multiplication is distributive, i.e., A(B + C) = AB + AC where A, B, C are any three m × n, n × p and n × p matrices respectively. Transpose of matrix Let A be any matrix. The matrix obtained by interchanging rows and columns of A is called the transpose of A and is denoted by A’ or At. Thus if A = [aij] is an m × n matrix, then At = [aji] will be n × m matrix. For example, the transpose of the matrix A= [ 2 3 1 2 4 0 ] 2×3 is 2 1 At = 3 2 4 0 [ ] 3×2 120 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Properties of transpose of matrices 1. Transpose of a sum (or difference) of two matrices is the sum (or difference) of the transposes, i.e., (A ± B)t = At ± Bt. 2. Transpose of a transpose is the original matrix (At)t = A. 3. Transpose of a product of two matrices is the product of their transposes taken in reverse order (AB)t = Bt At. Activity E If two matrices A and B are defined as A= [ 2 2 1 2 ; 4 0 ] 2 2 B= 1 4 2 0 [ ] then verify that (AB)t = Bt At. 4.5 Determinant of a square matrix The determinant of a square matrix is a scalar (i.e., a number). Determinants are possible only for square matrices. For more clarity, we shall be defining it in stages, starting with square matrix of order 1, then for matrix of order 2, etc. The determinant of a square matrix A is denoted either by |A| or det. A. 1. Determinant of order 1. Let A = (a11) be a matrix of order 1. Then det. A = a11. 2. Determinant of order 2. Let A= [ a11 a21 a12 a22 ] UNIT 1 121 Basic mathematics for management be a square matrix of order 2, then the determinant of A is defined as det. A = | a11 a12 = a11 a22 − a21 a12 a21 a22 | For example, det. A = | | 3 4 = 3 × 2 − 1 × 4 = 2. 1 2 To write the expansion of a determinant to matrices of order 3, 4, ..., let us first define two important terms: Minor: Let A be a square matrix of order m. Then the minor of an element aij is the determinant of the residual matrix (or submatrix) obtained from A by deleting row i and column j containing the element aij. In the matrix A, [ a11 a12 a21 a22 a31 a32 a13 a23 a33 ] | | The minor of the element aij is denoted by Mij. The minor of the element a11 is obtained by deleting the first row and first column containing element a11 and is written as M11 = | | a22 a23 . a32 a33 Similarly, minor of a12 is a21 a23 . a31 a33 M12 = Cofactor: The cofactor cij of an element aij is defined as cij = (−1)i+jMij where Mij is the minor of an element aij. 122 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Now using the concept of minor and cofactor, you can write the expansion of a determinant of order 3 as shown below: | a11 a21 a31 a12 a22 a32 a13 a23 a33 | = a11 C11 + a12 C12 + a13 C13 = a11 (−1)1+1 M11 + a12 (−1)1+2 M12 + a13(−1)1+3 M13. = a11 | a11 a32 a22 a − a12 21 a33 a31 | | a23 a + a13 21 a31 a33 | | a22 a32 | = a11(a22 a33 − a32 a23) − a12(a21 a33 − a31 a23) + a13(a21 a32 − a31 a22). The expansion of the given determinant can also be done by choosing the elements of any row and column. In the above example, expansion was done by using the elements of the first row. Example 4 Find the value of the determinant, 1 det. A = 2 2 | 18 40 45 72 96 . 75 | Solution: If you expand the determinant by using the elements of the first column, then you will get | 1 18 72 40 2 40 96 = 1 45 2 45 75 | | 96 18 −2 75 45 | | 72 18 72 +2 75 40 96 | | | = 1(3000 − 4320) − 2(1350 − 3240) + 2(1728 − 2880) = 1 × (−1320) − 2 × (−1890) + 2(−1152) = −1320 + 3780 − 2304 = −3624 + 3780 = 156. UNIT 1 123 Basic mathematics for management Properties of determinants The following are useful properties of determinants of any order. These properties are very useful in expanding the determinants. 1. The value of a determinant remains unchanged if rows are changed into columns and columns into rows, i.e., |A| = |At|. 2. If two rows (or columns) of a determinant are interchanged, then the value of the determinant obtained is the negative of the original determinant. 3. If each element in any row or column of a determinant is multiplied by a constant number, say K, then the determinant obtained is K times the original determinant. 4. The value of a determinant in which two rows (or columns) are equal is zero. 5. If any row or column of a determinant is replaced by the sum of the row and a linear combination of other rows (or columns), then the value of the determinant so obtained is equal to the value of the original determinant. 6. The rows (or columns) of a determinant are said to be linearly dependent if |A| = 0, otherwise they are independent. Example 5 Verify the following result. | | 1 a a2 1 b b2 = (a − b) (b − c) (c − a). 1 c c2 Applying row operations (Property 5), R2 o R2 + (−1)R1 R3 o R3 + (−1)R1 The determinant obtained is: | 1 0 0 a a2 b − a b2 − a2 c − a c2 − a2 | 124 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Expanding the new determinant by the elements of first column, you will get: | b−a c−a b2 − a2 b−a = c−a c2 − a2 || (b − a)(b + a) . (c − a)(c + a) | By performing row operations below: R2 o 1 R (b − a) 1 R (c − a) R3 o You will have (b − a)(c − a) | 1 1 b+a c+a | = (b − a)(c − a){(c + a) − (b + a)} = (b − a)(c − a)(c − b) = (a − b)(b − c)(c − a) Activity F If a + b + c = 0, then verify the following result. | | a b c 0 a b = c(2ab − c2) b 0 a 4.6 Inverse of a matrix For a given square matrix A, if another square matrix B of the same order fulfils the condition AB = BA = I then matrix B is called the inverse of A and is denoted by B = A−1. UNIT 1 125 Basic mathematics for management Before discussing the procedure of finding the inverse of a matrix, it is important to know the following results: 1. The matrix B = A−1 is said to be the inverse of matrix A if and only if AA−1 = A−1 A = I. 2. That is, if the inverse of a square matrix is multiplied by the original matrix, 1 I then the result is an identity matrix. The inverse A−1 does not mean or . A A This is simply a notation to denote the inverse of A. 3. Every square matrix may not have an inverse. For example, zero matrix has no inverse because the inverse of square matrix exists only if the value of its determinant is non-zero, i.e., A−1 exists if and only if |A| ≠ 0. For example, let B be the inverse of the matrix A, then AB = BA = I or |AB| = I or |A|.|B| = 1(|I| =1). Hence |A| ≠ 0. 4. If a square matrix A has an inverse, then it is unique. It can also be proven by assuming there are two inverses B and C of A. We then have AB = BA = I and AC = CA = I (ii) (i) 126 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Pre-multiplying (i) by C, we get CAB = CI or IB = CI or B = C(as CA = I). This implies that the inverse of a square matrix is unique. Singular matrix A matrix is said to be singular if its determinant is equal to zero; otherwise it is non-singular. Properties of the inverse 1. The inverse of the inverse is the original matrix, i.e., (A−1) = A. 2. The inverse of the transpose of a matrix is the transpose of its inverse, i.e., (At)−1 = (A−1)t. 3. The identity matrix is its own inverse, i.e., I−1 = I. 4. The inverse of the product of two non-singular matrices is equal to the product of two inverses in the reverse order, i.e., (AB)−1 = B−1. A−1. -1 Method of finding inverse of a matrix The procedure of finding inverse of a square matrix A = [aij] of order n can be summarised in the following steps: 1. Construct the matrix of co-factors of each element aij in |A| as follows: [ C11 C21 : : Cm1 C12 C22 : : Cm2 … … … C1n C2n : : Cmn ] In this case cofactors are the elements of the matrix. UNIT 1 127 Basic mathematics for management 2. Take the transpose of the matrix of cofactors constructed in step 1. It is called adjoint of A and is denoted by Adj. A. 3. Find the value of |A|. 4. Apply the following formula to calculate the inverse of A. A−1 = Adj A , where |A| ≠ 0. |A| Example 6 Find the inverse of the matrix: 1 A = −2 1 [ ] 3 3 1 0 3 4 Solution: The determinant of matrix A is expanded with respect to the elements of the first row: 1 |A| = −2 1 | | 3 3 1 0 3 3 = I 1 4 | | | 3 −2 −3 4 1 3 −2 +0 4 1 | | | 3 1 = 9 − 3(−11) = 42. Since |A| ≠ 0, therefore the inverse of A exists. The matrix of cofactor of elements A is: C11 = (−1)1+1 M11 = 1 C12 = (−1)1+2 M12 = −1 C13 = (−1)1+3 M13 = 1 C21 = (−1)2+1 M21 = −1 C22 = (−1)2+2 M22 = 1 | | | | | | | | | | 3 1 −2 1 −2 1 3 1 3 = 9. 4 3 = 11. 4 3 = −5. 1 0 = −12. 4 1 0 = 4. 1 4 128 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques C23 = (−1)2+3 M23 = −1 C31 = (−1)3+1 M31 = 1 C32 = (−1)3+2 M32 = −1 C33 = (−1)3+3 M33 = 1 | | | | | | | | 1 1 3 3 1 −2 3 = 2. 1 0 = 9. 3 1 0 = −3. −2 3 3 = 9. 3 The matrix of cofactor of A is [ Hence C11 C12 C13 9 11 C21 C22 C23 = −12 4 C31 C32 C33 9 −3 ][ [ −5 2 . 9 ] The adj. A is now constructed by taking the transpose of the cofactor matrix: 9 Adj. A = (Co-factor A)t 11 −5 −12 4 2 9 −3 9 ] A−1 = Adj. A |A| 9 1 11 42 −5 = [ −12 4 2 9 −3 9 ] Activity G For the matrix 1 A = −1 0 [ ] 4 2 0 0 0 2 a) Calculate A−1. b) Verify (At)−1 = (A−1)t. c) Verify (Adj. A)−1 = Adj. (A−1). UNIT 1 129 Basic mathematics for management 4.7 Solution of linear simultaneous equations As mentioned earlier in the previous section, matrix algebra is useful in solving a set of linear simultaneous equations involving more than two variables. Now the procedure for getting the solution will be demonstrated. Consider the set of linear simultaneous equations 2x + 5y − 2z = 3; x + y − z = 5 and 3x − 2y + z = 10. These equations can also be solved by using ordinary algebra. However, to demonstrate the use of matrix algebra, the first step is to write the given system of equations in matrix form as follows: AX = B where A is known as the coefficient matrix in which coefficients of x are written in the first column, coefficients of y in the second column and the coefficients of z in the third column. X is the matrix of unknown variables x, y and z, and B is the matrix formed with the right hand terms in equations which do not involve unknowns x, y and z. Generalising the situation, let us consider m linear equations in n unknowns x1, x2, …, xn, a11 x1 + a12 x2 + … a1n xn = b1 a21 x1 + a22 x2 + … a2n xn = b2 …………………………… am1 x1 + am2 x2 + … amn xn = bm Writing this system of equations in matrix form, AX = B where a11 a12 … a1n a a22 … a2n A = 21 …………………… am1 am2 … amn [ ] m×n 130 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques x1 x2 X= : : xn [] n×1 B= [] b1 b2 : : : bm m×1 Classification of linear equations If matrix B is a zero matrix, i.e., B = 0, then the system AX = 0 is said to be a homogeneous system. Otherwise, the system is said to be non-homogeneous. Homogeneous linear equations When the system is homogeneous, i.e., b1 = b2 = … = bm = 0, the only possible solution is X = 0 or x1 = x2 =…..xn = 0. It is called a trivial solution. Any other solution, if it exists, is called a non-trivial solution of the homogeneous linear equations. In order to solve the equation AX = 0, we perform an elementary operation or transformation on the given coefficient matrix A which does not change the order of the matrix. An elementary operation is any one of the following three types: 1. The interchange of any two rows (or columns). 2. The multiplication (or division) of the elements of any row (or column) by any non-zero number, e.g., the Ri (row i) can be replaced by KRi(K ≠ 0). 3. The addition of the elements of any row (or column) to the corresponding elements of any other row (or column) multiplied by any number, e.g., R i (row i) can be replaced by Ri + KR j where R j is the row j and K ≠ 0. The elementary operation is called row operation if it applies to rows and column operation if it applies to columns. UNIT 1 131 Basic mathematics for management For the purpose of applying these elementary operations, we form another matrix called augmented matrix as shown below: a11 a12 … a1n . b1 a21 a22 … a2n . b2 [A:B] = …………………… am1 am2 … amn . bm [ ] Solution method We shall apply Gauss-Jordon method (also called Triangular Form Reduction method) to solve homogeneous linear equations. In this method, the given system of linear equations is reduced to an equivalent simpler system (i.e., system having the same solution as the given one). The new system looks like: x 1 + b 1x 2 + c 1x 3 = d 1 x 2 + c 2x 3 = d 2 x 3 = d 3. This method helps, not only to find solution to homogeneous equations but also to non-homogeneous system of equations having any number of unknowns. Example 7 Solve the following system of equations using Gauss Jordon method. x1 + 3x2 − x3 = 0 2x1 − x2 + 4x3 = 0 x1 − 11x2 + 14x3 = 0 Solution: The given system of equations in matrix form is: [ 1 2 1 3 −1 −11 −2 4 14 ][ ] [ ] x1 0 x2 = 0 or AX = 0. x3 0 The augmented matrix becomes 1 3 [A:O] = 2 −1 1 −11 [ −2 : 0 4 :0 . 14 : 0 ] 132 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Applying elementary row operations, R2 o R2 − 2R1 R3 o R3 − R1 The new equivalent matrix is: [ [ 1 0 0 3 −2 : 0 7 8 :0 . −14 16 : 0 ] Again applying R3 o R3 − 2R2, the new equivalent matrix is: 1 0 0 3 −7 0 −2 : 0 8 :0 . 0 :0 ] The equations equivalent to the given system of equations obtained by elementary row operations are: x1 + 3x2 − 2x3 = 0 8 −7x2 + 8x3 = 0 or x2 − x3 = 0. 7 The last equation, though true, is redundant and the system is equivalent to x1 + 3x2 − 2x3 = 0 8 x2 − x3 = 0 7 This is not in triangular form because the number of equations is less than the number of unknowns. This system can be solved in terms of x3 by assigning an arbitrary constant value, k to it. The general solution to the given system is given by Let x3 = k, k  ƒ (any real number), Then x2 = 8k 7 x1 = 2k − 3 () 8k −10k = 7 7 UNIT 1 133 Basic mathematics for management Activity H Solve the following system of equations using Gauss-Jordon method. 1. 4x1 + x2 = 0 −8x1 + 2x2 = 0 x1 − 2x2 + 3x3 = 0 2x1 + 5x2 + 6x3 = 0 2. Non-homogeneous linear equations The non-homogeneous linear equations can be solved by any of the following three methods: 1. Matrix Inverse method. 2. Cramer’s method. 3. Gauss-Jordon method. Again, for the purpose of demonstrating the above solution methods, we shall consider three equations with three unknowns. 1. Matrix inverse method Let AX = B be the given system of linear equations, and A−1 be the inverse of A. Pre-multiplying both sides of the equation by A−1, A−1(AX) = A−1B (A−1A)X = A−1B IX = A−1B X = A−1B where I is the identity matrix. The value of X gives the general solution to the given set of simultaneous equations. This solution is thus obtained by (a) first finding A−1, and (b) post multiplying A−1 by B. When the system has a solution, it is said to be consistent, otherwise it is inconsistent. A consistent system has either just one solution or infinitely many solutions. 134 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Example 8 The daily cost, C of operating a hospital, is a linear function of the number of in-patients, I and out-patients, P, plus a fixed cost a, i.e., C = a + b P + dI. Given the following data for three days, find the values of a, b and d by setting up a linear system of equations and using the matrix inverse. Day (in RM) 1 2 3 Cost 6,950 6,725 7,100 No. of in-patients, I 40 35 40 No. of out-patients, P 10 9 12 Solution: Based on the given daily cost equation, the system of equations for the cost of three days can be written as: a + 10b + 40d = 6,950 a + 9b + 35d = 6,725 a + 12b + 40d = 7,100 This system can be written in the matrix form as follows: [ 1 10 40 1 9 35 1 12 40 ][ ] [ ] a 6,950 b = 6,725 c 7,100 which is of the form AX = B, where a 6,950 1 10 40 A= 1 9 35 ; X = b and B 6,725 . 1 12 40 c 7,100 [ ] [] [ ] ][ 5 0 −5 The inverse of a matrix A is obtained as follows: 60 C11 C12 C13 t Adj. A = C21 C22 C23 = −80 C31 C32 C33 10 [ −3 t 60 −80 10 2 = 5 0 −5 1 −3 2 1 ][ ] UNIT 1 135 Basic mathematics for management 1 |A| = 1 1 | 10 9 12 40 9 35 = 1 12 40 | | 35 1 − 10 40 1 | | 35 1 + 40 40 1 | | 9 12 | = (360 − 420) − 10(40 − 35) + 40(12 − 9) = −10 ≠ 0. Since |A| ≠ 0, therefore inverse of matrix A exists and is computed as Adj. A |A| 60 1 5 10 −3 A−1 = =− [ −80 0 2 10 −5 1 ] 10 −5 1 6,950 6,725 7,100 ?X = A−1B or [ ] [ ][ ] [ ] [ ][ ] a 60 1 b =− 5 10 −3 c −80 0 2 =− 60 × 6,950 − 80 × 6,725 + 10 × 7,100 1 5 × 6,950 + 0 × 6,725 − 5 × 7,100 10 −3 × 6,950 + 2 × 6,725 + 1 × 7,100 1 10 −50,000 5000 −750 = 75 −300 30 =− or a = 5000, b = 75 and d = 30. Activity I A salesman has the following record of sales during three months for three items A, B and C, which have different rates of commission. Sales of units A January February March 90 150 60 B 100 50 100 C 20 40 30 Total commision drawn (in RM) 800 900 850 Months 136 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Find out the rates of commission on items A,B and C. 2. Cramer’s method When the number of equations is equal to the number of unknowns and the determinant of the coefficients has non-zero value, then the system has a unique solution which can be found by using Cramer’s formula. xj = Dj , j = 1, 2, …, n D where D = |aij| and determinant Dj is obtained from D by replacing column j by the column of constant terms (i.e., matrix B). Example 9 An automobile company uses three types of steel, S1, S2 and S3 for producing three different types of cars C1, C2 and C3. Steel requirements (in tons) for each type of car and the total available steel of all the three types is summarised in the following table. Types of car C1 S1 S2 S3 2 1 3 C2 3 1 2 C3 4 2 1 29 13 16 Types of steel Total steel available Determine the number of cars of each type which can be produced. Solution: Let x1, x2 and x3 be the number of cars of the types C1, C2 and C3 respectively which can be produced. The system of three linear equations is: 2x1 + 3x2 + 4x3 = 29 x1 + x2 + 2x3 = 13 3x1 + 2x2 + x3 = 16. These equations can also be represented in matrix form as shown below: [ ][ ] [ ] 2 3 4 1 1 2 3 2 1 x1 29 x2 = 13 . x3 16 UNIT 1 137 Basic mathematics for management The determinant of the coefficients matrix is 2 3 4 1 D= 1 1 2 =2 2 3 2 1 | | | | | | | | | | | 2 1 −3 1 3 2 1 +4 1 3 1 2 = 2(1 − 4) − 3(1 − 6) + 4(2 − 3) = 5 ≠ 0. Applying Cramer’s method, 29 3 D1 1 = 13 1 D 5 16 2 2 29 D2 1 = 1 13 D 5 3 16 4 2 =2 1 4 2 =3 1 x1 = x2 = x3 = 2 3 29 D3 1 = 1 1 13 = 4. D 5 3 2 16 | | | Hence, the number of cars of types C1, C2 and C3 which can be produced are 2, 3 and 4 respectively. Activity J A firm makes two products A and B. Each product requires production time in each of the two departments I and II as shown below: Time taken (in hrs/week) Dept I A B 5 6 Dept II 4 2 Product Total time available is 80 hours and 60 hours in departments I and II respectively. Determine the number of units of products A and B which should be produced. 138 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques 4.8 Applications of matrices Markov models A particular mathematical model which is concerned with the brand-switching behaviour of consumers who are essentially repeat buyers of the product, is known as Markov brand-switching model. These models help in predicting the market share of a product at time period t, if the market share at the time period (t−1) is known. Markov models have also been used in the study of (1) equipment maintenance and failure probability (2) stock market price movements, etc. The general expression for forecasting the buying levels at time t = n + 1 is given by Rn+1 = Rnp P11 P12 … P1n P21 P22 … P2n where p = …………………… Pm1 Pm2 … Pmn [ ] n j is the matrix of transition probabilities. Each element of it represents the probability that a customer will change his liking from one brand to another in his next purchase. This is the reason for calling them transition probabilities and ∑Pij = 1, R = matrix of order (1 × n) representing the buying levels (or state probabilities) at a particular time period. If we know the buying levels at time t = 0, then we can find them at any time by solving the above equation by the relation: R1 = R0P, n = 0 R2 = R1P = (R0P) P = R0P2; n = 1 ………………………………… Now Rn = R0Pn; n = n − 1. Now as the time passes, i.e., n o ∞ the purchasing levels (or market shares) tend to settle down to an equilibrium (or steady state). That is, once an equilibrium state is reached, there will be no change in the future market shares. Thus no∞ Lt. Rn+1 = nLt. Rn . P ∞ o or R = RP. This relationship can be used to determine market shares in the long run. UNIT 1 139 Basic mathematics for management Example 10 Consider the following matrix of transition probabilities of a product available in the market in two brands: Brand A Brand A Brand B 0.9 0.3 Brand B 0.1 0.7 Determine the market shares of each of the brand in equilibrium position. Solution: If the row vector (matrix having only one row) represents the market share of the two brands at equilibrium, then R = RP i.e., (r1 r2) = (r1 r2) [ 0.9 0.3 0.1 0.7 ] (1) (2) Ÿ r1 = 0.9r1 + 0.3r2 r2 = 0.1r1 + 0.7r2 or or −0.1r1 + 0.3r2 = 0 0.1r1 − 0.3r2 = 0 These are two linear homogeneous simultaneous equations. But these are not independent since one can be derived from the other. Hence, in order to solve, one more equation is needed, which is r1 + r2 = 1 (3) This is because the market shares have been expressed in percentage, so the sum of market shares will be 1. Solving equations (1) and (2) with the help of equation (3), to get market shares in an equilibrium condition, r1 = 0.75 and r2 = 0.25. Hence the expected market shares in an equilibrium condition for brand A will be 0.75 and that of brand B will be 0.25. 140 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Activity K The purchase patterns of two brands of toothpaste can be expressed as a Markov process with the following transition probabilities: Formula A Formula A Formula B 0.90 0.05 Formula B 0.10 0.95 What are the projected market shares for the two formulas? Input-output analysis The method of “input-output analysis” was first proposed by Wassily W. Leotief in the 1930s. This method is based on the concept of “economic inter-dependence”, which means that every sector (or industry) of the economy is related to every other sector. That is, they are all inter-dependent and inter-related. This means, any change in one sector (such as strike) will affect all other industries to a varying degree. However, this technique does not explain or establish why such effects occur. The input-output model is based on the following assumptions: 1. An economy is decomposed into n sectors (or industries), and each of these produces only one kind of product. Each of the sectors uses the output of the other sectors as input. Let xj (j =1, 2, ..., n) be the gross production (output) of the jth sector. 2. Let aij represents RM value of the output from sector i which sector j must consume to produce one RM worth of its own product. It can be calculated as follows: RM value of the product of sector i required by sector j RM value of the total output of sector j aij = The aij’s for all i and j can be represented in matrix form as shown below: a11 a12 … a1n a21 a22 … a2n A= …………………… an1 an2 … ann [ ] n×n UNIT 1 141 Basic mathematics for management The matrix A is the technical input-output coefficient matrix. This matrix remains unchanged so long as the structure of the economy remains unchanged. 3. There are neither shortages nor surpluses of product under consideration. In other words, the gross product of each sector is sufficient to meet the final demand as well as demands of other sectors. Let dj (j =1, 2, ..., n) be the final demand (in RM value) for the product produced by each of n sectors. The input-output table displayed in the following table summarises information about the economy in question. Customer sectors Producers sectors Sector 1 2 x1 x2 xi xn Intermediate use for sectors 1 a11 a21 ai1 an1 2 a12 a22 ai2 an2 … … … j a1j a2j aij anj … … … n a1n a2n ain ann Final use x1 x2 xi xn Final demand d1 d2 di dn ………… i ……………………………… … … ………… n ……………………………… … … If the economy is assumed to be in a state of dynamic equilibrium (i.e. neither shortages nor surpluses) so that the total output is just sufficient to meet the input needs of each sector as well as the needs of the final demand of all sectors themselves, then Output = Input = Need of each sector + Final demand n xi = ∑ aijxj + di; for sector i = 1, 2, …, n. j=1 In matrix notation, we have X = AX + D x1 a11 a12 … a1n x2 a21 a22 … a2n where X ,A= and D = : …………………… an1 an2 … ann xn [] [ ] [] d1 d2 . : dn 142 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques The above equation can also be rewritten as: X = AX + D IX = AX + D IX − AX = D (I − A)X = D (I − A)X = D X = (I − A)−1 D provided |I − A| ≠ 0 where I is the identity matrix. The value of X gives how much each sector must produce which is just sufficient to meet the final demand as well as the demand of all sectors themselves. Example 11 Given the following input-output table, calculate the gross output so as to meet the final demand of 200 units of Agriculture and 800 units of Industry. Input value (in RM ‘000) Agriculture Agriculture Industry 300 400 Industry 600 1200 1000 2000 Producer Total output (in RM ‘000) Solution: Using the notations as discussed above, a11 = RM value of the product of sector Agriculture used by Agriculture RM value of the total output of sector Agriculture 300 = 0.3. 1000 = Similarly, a12 = 600 = 0.3. 2000 400 = 0.4. 1000 1200 = 0.6. 2000 a21 = a22 = Thus the technological matrix A and the final demand matrix D become UNIT 1 143 Basic mathematics for management A= [ 0.3 0.4 Now I − A = |I − A| = (I − A)−1 = X = (I − A)−1D = ] [ ] [ ][ ][ [ | | [ ] [ ][ ] [ ] [ ] 0.3 200 ;D= 0.6 800 1 0 0 0.3 − 1 0.4 = 0.7 −0.4 Adj. (I − A) 1 = |I − A| 0.16 0.4 0.3 0.4 0.7 200 800 1 0.4 0.3 0.16 0.4 0.7 1 320 0.16 640 2000 . 4000 = = 0.3 1 − 0.3 = 0.6 0 − 0.4 0.7 −0.4 0 − 0.3 1 − 0.6 −0.3 . 0.4 ] ] −0.3 = 0.28 − 0.12 = 0.16 ≠ 0 0.4 Hence, the gross output of Agriculture and Industry must be 2000 units and 4000 units respectively. Activity L In an economy there are two sectors A and B and the following table gives the supply and demand position of these in million RM: Consumer A A B 15 20 B 10 30 35 50 Producer Total output Determine the total output, if the demand is 12 for A and 18 for B. 144 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Summary Matrices play an important role in quantitative analysis of managerial decisions. They also provide very convenient and compact methods of writing a system of linear simultaneous equations and solving them. These tools have also become very useful in all functional areas of management. Another distinct advantage of matrices is that once the system of equations can be set up in matrix form, they can be solved quickly using a computer. A number of basic matrix operations (such as matrix addition, subtraction, multiplication) were discussed in this section. This was followed by a discussion on matrix inversion and the procedure for finding matrix inverse. A number of examples were given in support of the above said operations and inverse of a matrix. Finally, two important applications of matrix algebra predicting market shares using Markov models and predicting the effect of a change in the output (or demand) of one sector of the economy on the output of the other sectors, using input-output models were discussed. Keywords Co-factor: The number Cij = (−1)i+jMij is called the co-factor of element aij in A. Determinant: A unique scalar quantity associated with each square matrix. Identity matrix: A matrix in which diagonal elements are equal to 1 and all other elements are zero. Matrix: It is an array of numbers, arranged in rows and columns. Minor: The minor of an element is the determinant of the submatrix obtained from a given matrix by deleting the row and the column containing that element and is denoted by Mij. Null matrix: A matrix in which all elements are zero. Transpose matrix: A new matrix obtained by interchanging rows and columns of the original matrix. UNIT 1 145 Basic mathematics for management Further readings Tan, S T (2007) Applied Mathematics for the Managerial, Life and Social Sciences, 4th edn, Belmont, CA: Thomson Brooks/Cole. Haeussler, E F, Paul R S and Wood, R J (2005) Introductory Mathematical Analysis for Business, Economics and the Life and Social Sciences, 11th edn, Upper Saddle River, NJ: Pearson Prentice Hall. Lial, M L, Hungerford, T W and Holcomb, J P (2007) Mathematics with Applications in the Management, Natural and Social Sciences, 9th edn, Boston: Pearson Addison Wesley. Raghawachari, M (1985) Mathematics for Management: An Introduction, Delhi: Tata McGraw Hill (India). 146 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Suggested answers to activities Activity A 1. A) 3 × 3 Activity B a) b) c) [ [ [ 0 2 0 2 7 1 2 1 2 1 6 4 3 7 + 4 1 3 7 − 4 1 3 0 − 5 2 ][ ][ ][ 6 4 6 4 2 1 3 7 = 5 3 3 −7 = 5 1 3 7 = 4 −1 ][ ][ ][ 8 5 −4 −3 4 3 6 9 ] 0 −1 0 1 ] ] Activity C 2 [ 0 2 2 3 7 +3 1 4 1 ] [ 6 3 0 = 4 5 4 ][ 4 2 6 21 18 + 8 3 12 ][ 9 21 = 15 7 ][ 22 15 14 23 ] Activity D 1. A) m × p 2. C) May be a zero matrix. 3. a) The number of worker matrix = (50 20) b) Total daily payment is (50 20) () 30 = 1500 + 340 = 1840. 17 UNIT 1 147 Basic mathematics for management Activity E 2 2 2 1 2 At = 1 4 , Bt = 2 4 0 2 0 [ ] [ [ 2 1 2 4 [ ] ] BtAt = 2 2 2 1 2 9 8 1 4 = 2 4 0 8 20 2 0 2 2 2 9 8 1 4 = 0 8 20 2 9 AB = ][ ][ ] [ ] ][ ] (AB)t = [ 9 8 8 20 (AB)t = BtAt Activity F Given that a + b + c = 0. Thus a + b = −c. a | | | | | | a 0 b 0 b 0 a −b +c = a(a2) − b(−b2) −(−ab) a b 0 b 0 = a3 + b3 − abc = (a + b)3 − 3a2b − 3ab2 − abc = −c3 − 3ab(a + b) − abc = −c3 − 3ab(−c) − abc = −c3 + 3abc − abc = c(2ab − c2) 148 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques Activity G a) |A| = 12 4 Adj. A = 2 0 4 1 2 12 0 A −1 = [ [ −8 2 0 −8 2 0 0 0 6 0 0 6 ] ] b) (A −1) = 4 2 0 1 −8 2 0 12 0 0 6 [ ] 1 At = 4 0 |At| = 12 [ −1 0 2 0 0 2 ] (A t) = −1 4 2 0 1 −8 2 0 12 0 0 6 [ ] 12 1 −12 144 0 ? (A −1)t = (A t ) −1 c) |Adj. A| = 144. (Adj. A) −1 = [ 48 24 0 0 0 = 1 − 24 12 0 ] Adj. (A −1) = [ ] 1 12 4 12 0 − 1 12 0 2 12 0 0 1 12 [ ] 1 12 4 0 12 2 12 0 1 12 0 . ?(Adj. A)−1 = Adj.(A−1) UNIT 1 149 Basic mathematics for management Activity H 1. [ |] [ | ] () [ | ] [ | ] () [ | ] 4 1 0 R2 o 2R1 + R2 −8 2 0 4 0 0 1 0 0 1 R1 o R1 0 1 0 4 0 1 0 x2 = 0; x1 = 0 4 1 0 4 1 0 1 R2 o R2 R1 = (−1)R2 + R1 0 4 0 4 0 1 0 2. [ [ [ 1 2 −2 3 0 R2 o 2R1 − R3 5 6 0 1 −2 3 0 1 1 R2 o − R2 0 −9 0 0 9 0 1 0 3 0 0 1 0 0 x2 = 0 x3 = k, k  R |] |] ( ) [ |] −2 3 0 R1 o (2)R2 + R1 1 0 0 |] x1 + 3x3 = 0 x1 = −3k Activity I Let a = rate of commission on item A b = rate of commission on item B c = rate of commission on item C [ ][ ] [ ] [] [ ] [ ] [] [ ][ ] 90 150 60 100 50 100 20 40 30 a 800 b = 900 c 850 20 40 30 −1 a 90 b = 150 c 60 100 50 100 800 900 850 −2.49 × 10−3 a b = −2.09 c 11.94 8.96 × 10−3 −4.48 −2.99 −1.99 14.33 10.45 800 900 850 150 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques 4.39 × 10−3 = 6.48 × 10−3 15.74 × 10−3 [ ] Activity J Let x = No. of units of product A produced y = No. of units of product B produced 5x + 6y = 80 6x + 2y = 60 [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [] 5 6 6 2 x 80 = y 60 1 2 x = y −26 −6 = −6 80 5 60 1 −200 −26 −180 = 7.69 8 ≈ 6.92 7 Activity K (r1, r2) = (r1, r2) [ 0.9 0.05 0.1 0.95 ] r1 = 0.9r1 + 0.05r2 0.1r1 − 0.05r2 = 0 (1) r2 = 0.1r1 + 0.95r2 0.1r1 − 0.05r2 = 0 (2) r1 + r2 = 1 r2 = 1 − r1 (3) o (1) (3) 0.1r1 − 0.05(1 − r1) = 0 0.15r1 = 0.05 r1 = 0.33 r2 = 1 − 0.33 = 0.67. UNIT 1 151 Basic mathematics for management Activity L A= I−A= [ ] [ ] 15 35 10 50 20 30 35 50 ;D= 20 35 10 − 50 20 50 − 20 35 X = (I − A)−1D [ ] 12 18 |I − A| = 400 − 200 4 = ≠0 35(50) 35 20 10 35 50 50 (I − A)−1 = 4 20 20 35 35 20 10 35 50 50 = 4 20 20 35 35 [ ] [ ] [ ] 12 18 [][ ] A 73.5 = B 150 The total output for products A and B are 73.5 and 150 respectively. 152 WAWASAN OPEN UNIVERSITY BBM 511 Quantitative Techniques