SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT BUSINESS MATHEMATICS / MATHEMATICAL ANALYSIS

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1 SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT BUSINESS MATHEMATICS / MATHEMATICAL ANALYSIS Unit Six Moses Mwale moses.mwale@ictar.ac.zm

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3 BBA 120 Business Mathematics Contents Unit 6: Matrix Algebra Matrices Basic Definitions Special Matrices... 4 Example Operations with Matrices... 5 Transpose... 5 Sum, Difference... 6 Scalar Multiple... 6 Example: Demand Vectors for an Economy... 6 Matrix Multiplication Matrix Equations... 9 Example:... 9 Exercises: Perform the indicated operations... 9 Matrices and Linear Systems Solving Systems of Equations Determinants Example The Inverse of a Matrix Why Would We Want an Inverse? A Real Life Example Order is Important Bigger Matrices Inverse of a Matrix using Minors, Cofactors and Adjugate More Examples Cramer's Rule Exercises:... 26

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5 BBA 120 Business Mathematics Unit 6: Matrix Algebra 6.0 Matrices 6.1 Basic Definitions ROWS ARE HORIZONTAL AND COLUMNS ARE VERTICAL. An m n matrix is a rectangular array of real numbers with m rows and n columns. The numbers m and n are the dimensions of A. The real numbers in the matrix are called its entries. The entry in row i and column j is called a ij or A ij. To denote the entries in a matrix A of size n x m, say, we use the name of the matrix, with double subscripts to indicate position, consistent with the conventions above: For the entry A l2 (read "A sub onetwo" or just "A one-two"), the first subscript, 1, specifies the row and the second subscript, 2, the column in which the entry appears. Similarly, the entry A 23 (read "A two-three") is the entry in the second row and the third column. Generalizing, we say that the symbol A ij denotes the entry in the i th row and j th column. In fact, a matrix A is a function of two variables with A(i,j) = A ij. Example Following is a 4 5 matrix with the entry A 23 highlighted A = [ ] A matrix that has exactly one row, such as the 1 4 matrix is called a row vector. A = [ ] A matrix consisting of a single column; such as the 5 1 matrix 3

6 4 Unit 6: Matrix Algebra is called a column vector [ 16] 6.2 Special Matrices Zero matrix Certain types of matrices play important roles in matrix theory. We now consider some of these special types. An m x n matrix whose entries are all 0 is called the m n zero matrix and is denoted by O m n or, more simply, by 0 if its size is understood. Thus, the 2 x 3 zero matrix is Square matrix Identity matrix O = [ ] A matrix having the same number of columns as rows-for example, n rows and n columns-is called a square matrix of order n. That is, an m x n matrix is square if and only if m = n. For example, matrices [ ] and [5] are square with orders 3 and,1, respectively. Just as the zero matrix plays an important role as the identity in matrix addition there is a special matrix, called the identity matrix that plays a corresponding role in matrix multiplication: The n x n identity matrix, denoted I is the diagonal matrix whose main diagonal entries are 1's. For example, the identity matrices I 3 and I 4 are I 3 = [ 0 1 0] I 4 = [ ]

7 BBA 120 Business Mathematics Example Solution A grocer sold 125 cans of tomato soup, 275 cans of beans, and 400 cans of tuna. Write a row vector that gives the number of each item sold. If the items sell for $0.95, $1.03, and $1.25 each, respectively, write this information as a column vector. Row Vector Column vector Tomato Soup, Beans, Tuna [ ] 0.95 [ 1.03] Operations with Matrices Transpose The transpose, A T, of a matrix A is the matrix obtained from A by writing its rows as columns. If A is an m n matrix and B = A T, then B is the n m matrix with b ij = a ji. Examples 5

8 6 Unit 6: Matrix Algebra [ ] 3 T 0 = [ ] Sum, Difference If A and B have the same dimensions, then their sum, A+B, is obtained by adding corresponding entries. In symbols, (A+B) ij = A ij + B ij. If A and B have the same dimensions, then their difference, A - B, is obtained by subtracting corresponding entries. In symbols, (A-B) ij = A ij - B ij. Properties of Matrix Addition 1. A+B=B+A commutative property 2. A + (B + C) = (A + B) + C associative property 3. A+O =A= O+A identity property Scalar Multiple If A is a matrix and c is a number (sometimes called a scalar in this context), then the scalar multiple, ca, is obtained by multiplying every entry in A by c. In symbols, (ca) ij = c(a ij). Example [ 1 3 1] + 2 [ 2 3 2] = [ 5 3 5] Properties of Scalar Multiplication 1. k(a + B) = ka + kb. 2. (k + l)a = ka + la 3. k(la) = (kl)a 4. 0A = A 5. k0 = 0 We also have the following properties of the transpose operation, where A and B are of the same size and k is any scalar: (A + B) T = A T + B T (ka) T = ka T Example: Demand Vectors for an Economy Let matrix A represent the sales (in thousands of dollars) of a toy company in 2007 in three cities, and let B represent the sales in the same cities in 2009, where

9 BBA 120 Business Mathematics A = Action Educational [ ] Solution B = Action Educational [ ] If the company buys a competitor and doubles its 2009 sales in 2010, what is the change in sales between 2007 and 2010? Let the sales in 2010 be denoted by matrix C Then C = 2B = 2 [ ] = [ ] The change in sales between 2007 and 2012 is given by C A = [ ] [ ] = [ ] Matrix Multiplication If A has dimensions m n and B has dimensions n p, then the product AB is defined, and has dimensions m p. The entry (AB) ij is obtained by multiplying row i of A by column j of B, which is done by multiplying corresponding entries together and then adding the results. Three points must be completely understood concerning this definition of AB. First, the number of columns of A must be equal to the number of rows of B. Second, the product AB has as many rows as A and as many columns as B. Third, the definition refers to the product AB, in that order; A is the left factor and B is the right factor. For AB, we say that B is pre-multiplied by A or A is post-multiplied by B. Examples 7

10 8 Unit 6: Matrix Algebra 1. Let A be a 3 x 5 matrix and B be a 5 x 3 matrix. Then AB is defined and is a 3 x 3 matrix. Moreover, BA is also defined and is a 5 x 5 matrix. 2. If C is a 3 x 5 matrix and D is a 7 x 3 matrix, then CD is undefined, but DC is defined and is a 7 x 5 matrix. 3. [ ] [ ] = [ 4. Find a, b and c so that ] i) [ 1 3 b 5 ] [a ] = [6 1 4 c d 7 7 ] ii) [ 1 2 b ] [a 2 3 c d ] = [ ] 5. Given the following matrices A = [ ] B = [ ] C = [ 4 3 1] D = [ 0 1] Perform the indicated operations, if possible. i. CA vii. C+DA ii. AC viii. B+AD iii. BA ix. 0.2CD iv. AB x. ( 2)BA + 6CD v. C 2 xi. CDA vi. B 2 xii. BAD Suppose that the prices (in dollars per unit) for products A, B, and C are represented by the price vector Price of A B C P = [2 3 4] If the quantities (in units) of A, B, and C that are purchased are given by the column vector 7 Units of A Q = [ 5 ] Units of B 11 Units of C then the total cost (in dollars) of the purchases is given by the entry in the cost vector 7 PQ = [2 3 4] [ 5 ] = [(2 7) + (3 5) + (4 11)] = [73] 11

11 BBA 120 Business Mathematics 6.4 Matrix Equations Example: Systems of linear equations can be represented by using matrix multiplication. For example, consider the matrix equation [ x ] [ x 2 ] = [ 4 x 3 ] (1) 3 The product on the left side has order 2 x 1 and hence is a column matrix. Thus, [ x 1 + 4x 2 2x 3 2x 1 3x 2 + x 3 ] = [ 4 3 ] By equality of matrices, corresponding entries must be equal, so we obtain the system x 1 + 4x 2 2x 3 = 4 2x 1 3x 2 + x 3 = 3 Hence, this system of linear equations can be defined by matrix Equation (1). We usually describe Equation (1) by saying that it has the form AX = B where A is the matrix obtained from the coefficients of the variables, X is a column matrix obtained from the variables, and B is a column matrix obtained from the constants. Matrix A is called the coefficient matrix for the system. Write the system { 2x 1 + 5x 2 = 4 8x 1 + 3x 2 = 7 in matrix form by using matrix multiplication. Solution A = [ ] X = [x 1 x 2 ] B = [ 4 7 ] then the given system is equivalent to the single matrix equation AX = B [ ] [x 1 x 2 ] = [ 4 7 ] Exercises: Perform the indicated operations [ ] [ ]

12 10 Unit 6: Matrix Algebra 2. [ ] ([ ] + [ ]) x 3. [ 0 1 0] [ y] z 4. [ a 11 a 12 a 21 a 22 ] [ x 1 x 2 ] 5. A stockbroker sold a customer 200 shares of stock A, 300 shares of stock B, 500 shares of stock C, and 25.0 shares of stock D. The prices per share of A, B, C, and Dare $100, $150, $200, and $300, respectively. Write a row vector representing the number of shares of each stock bought. Write a column vector representing the price per share of each stock. Using matrix multiplication, find the total cost of the stocks. Matrices and Linear Systems A linear system is the system of linear equations a 11 x 1 + a 12 x a 1n x n = b 1 ; a 21 x 1 + a 22 x a 2n x n = b 2 ; (6.0.1) a m1 x 1 + a m2 x a mn x n = b m : Solving the linear system (6.0.1) means finding the variables x 1, x 2,, x n that satisfy all the equations in (6.0.1) simultaneously. The connection between linear systems and matrices is an obvious one. Indeed, let a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n x 2 b A=[ ], x= [ ] and b= [ 2 ] a m1 a m2 a mn x n b n then the linear system (6.0.1) may be rewritten as Ax = b. 6.5 Solving Systems of Equations

13 BBA 120 Business Mathematics In this section, we ll study one method for solving systems of linear equations using matrices called Crammer s Rule. This method works only for systems with the same number of equations and variables. The biggest advantage of Cramers is that it s purely computational it requires very little symbol manipulation. The method is based on determinants Determinants For any square matrix A, the determinant of A is a real number denoted by det(a) or A. If A is a square matrix of order n, then det (A) is called a determinant of order n. If A = [a 11 ] is a square matrix of order 1, then det(a) = a 11 is a first-order determinant. Now we proceed to define determinants of higher order. Given a second-order square matrix A = [ a 11 a 12 a 21 a ], the second-order 22 determinant of A is det(a) = a 11 a 12 a 21 a 22 = a 11 a 22 a 21 a 12 Formula above is easily remembered if you notice that the expression on the right is the product of the elements on the principal diagonal, from upper left to lower right, minus the product of the elements on the secondary diagonal, from lower left to upper right. Example Evaluating the Second Order Determinant. Find Solutions det(a) = = ( 1)( 4) ( 3)(2) = 4 ( 6) = 10 Evaluating Third-Order Determinants a 11 a 12 a 13 Given the matrix A = [ a 21 a 13 a 22 a 32 a 23 ] the third-order determinant of A a 33 is 11

14 12 Unit 6: Matrix Algebra a 11 a 12 a 13 det(a) = a 21 a 13 a 22 a 32 a 23 a 33 = a 11 a 22 a 33 a 11 a 32 a 23 + a 21 a 32 a 13 a 21 a 12 a 33 + a 31 a 12 a 23 a 31 a 22 a 13 Caution Remember that A = [ ] is a matrix, but represents a real number, the determinant of A. We will often refer to as a determinant, and refer to the process of finding the real number it represents as evaluating the determinant. Don t panic! You don t need to memorize this formula. There is another method of finding the determinant of a 3 3 matrix. To find the determinant of a 3 3 matrix, copy the first two columns of the matrix to the right of the original matrix. Next, multiply the numbers on the three downward diagonals, and add these products together. Multiply the numbers on the upward diagonals, and add these products together. Then subtract the sum of the products of the upward diagonals from the sum of the product of the downward diagonals (subtract the second number from the first number): a 1 b 1 c 1 A = [ a 2 b 2 c 2 ] a 3 b 3 c 3 a 1 b 1 c 1 [ a 2 b 2 c 2 ] a 3 b 3 c 3 a 1 b 1 a 2 b 2 a 3 b Example: Find the determinant of: [ ] Solution: Step [ ]

15 BBA 120 Business Mathematics Step 2 Step 3 Step = -70. deta =

16 14 Unit 6: Matrix Algebra Reciprocal of a Number But we don't write 1 / A (because we don't divide by a Matrix!), instead we write A -1 for the inverse: The Inverse of a Matrix. What is the Inverse of a Matrix? The Inverse of a Matrix is the same idea as the reciprocal of a number: When you multiply a number by its reciprocal you get 1 8 ( 1 / 8) = 1 When you multiply a Matrix by its Inverse you get the Identity Matrix (which is like "1" for Matrices): A A 1 = I It also works when the inverse comes first: ( 1 / 8) 8 = 1 and A -1 A = I Note: the "Identity Matrix" is the matrix equivalent of the number "1": Definition The Inverse of A is A -1 only when: Sometimes there is no Inverse at all. A A 1 = A 1 A = I How to calculate the Inverse of a matrix. Well, for a 2x2 Matrix the Inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad bc). Let us try an example:

17 BBA 120 Business Mathematics How do we know this is the right answer? Remember it must be true that: A A -1 = I So, let us check to see what happens when we multiply the matrix by its inverse: We end up with the Identity Matrix! So it must be right. It should also be true that: A -1 A = I Why Would We Want an Inverse? Because with Matrices we don't divide! There is no concept of dividing by a Matrix. But we can multiply by an Inverse, which achieves the same thing. Say that you know Matrix A and B, and want to find Matrix X: XA = B It would be nice to divide both sides by A (to get X=B/A), but remember we can't divide. But what if we multiply both sides by A -1? And we know that AA -1 = I, so: XAA 1 = BA 1 XI = BA 1 We can remove I (for the same reason we could remove "1" from 1x = ab for numbers): X = BA 1 And we have our answer (assuming we can calculate A -1 ) 15

18 16 Unit 6: Matrix Algebra In that example we were very careful to get the multiplications correct, because with Matrices the order of multiplication matters. AB is almost never equal to BA. A Real Life Example A group travelled on a rented bus, at K3 per child and K3.20 per adult for a total of K They took the train back at K3.50 per child and K3.60 per adult for a total of K How many children, and how many adults were in the group? First, let us set up the matrices (be careful to get the rows and columns correct!): This is just like the example above: XA = B So to solve it we need the inverse of "A": Now we have the inverse we can solve using: X = BA 1 There were 16 children and 22 adults!

19 BBA 120 Business Mathematics The answer almost appears like magic. But it is based on good mathematics. It is also a way to solve Systems of Linear Equations. The calculations are mostly done by computer, but the people must understand the formulas. Order is Important Say that you are trying to find "X" in this case: AX = B This is different to the example above! X is now after A. With Matrices the order of multiplication usually changes the answer. Do not assume that AB = BA, it is almost never true. So how do we solve this one? Using the same method, but put A -1 in front: And we know that A 1 A = I, so: We can remove I: A 1 AX = A 1 B IX = A 1 B X = A 1 B And we have our answer (assuming we can calculate A -1 ) Why don't we try our example from above, but with the data set up this way around. (Yes, you can do this, just be careful how you set it up.) This is what it looks like as AX = B: It looks so neat! I think I prefer it like this. Also note how the rows and columns are swapped over ("Transposed") compared to the previous example. To solve it we need the inverse of "A": 17

20 18 Unit 6: Matrix Algebra It is like the Inverse we got before, but Transposed (rows and columns swapped over). Now we can solve using: X = A -1 B Points to Remember The Inverse of A is A -1 only when A A -1 = A -1 A = I Same answer: 16 children and 22 adults. So, Matrices are powerful things, but they do need to be set up correctly! To find the Inverse of a 2x2 Matrix: swap the positions The Inverse May Not Exist of a and d, put First of all, to have an Inverse the Matrix must be "Square" (same number negatives in front of rows and columns). of b and c, and But also the determinant cannot be zero (or you would end up dividing divide everything by zero). How about this: by the determinant. Sometimes there is no Inverse at all 24-24? That equals 0, and 1/0 is undefined. We cannot go any further! This Matrix has no Inverse. Such a Matrix is called "Singular", which only happens when the determinant is zero. And it makes sense... look at the numbers: the second row is just double the first row, and does not add any new information. Imagine in our example above that the prices on the train were exactly, say, 50% higher... we wouldn't be any closer to figuring out how many adults and children... we need something different. And the determinant neatly works this out. Bigger Matrices The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). For those larger matrices there are three main methods to work out the inverse:

21 BBA 120 Business Mathematics Inverse of a Matrix using Elementary Row Operations (Gauss- Jordan) Inverse of a Matrix using Minors, Cofactors and Adjugate Use a computer (such as the Matrix Calculator) Inverse of a Matrix using Minors, Cofactors and Adjugate You can calculate the Inverse of a Matrix by 1) calculating the Matrix of Minors, 2) then turn that into the Matrix of Cofactors, 3) then the Adjugate, and 4) multiply that by 1/Determinant. But it is best explained by working through an example! Example: find the Inverse of A: It needs 4 steps. It is all simple arithmetic but there is a lot of it, so try not to make a mistake! 19

22 20 Unit 6: Matrix Algebra A minor of a matrix A is the determinant of some smaller square matrix, cut down from A by removing one or more of its rows or columns. Minors obtained by removing just one row and one column from square matrices (first minors) are required for calculating matrix cofactors, which in turn are useful for computing both the determinant and inverse of square matrices. Step 1: Matrix of Minors The first step is to create a "Matrix of Minors": For each element of the matrix: ignore the values on the current row and column calculate the determinant of the remaining values Put those determinants into a matrix (the "Matrix of Minors") Here are the first two, and last two, calculations of the "Matrix of Minors" (notice how I ignore the values in the current row and columns, and calculate the determinant using the remaining values): And here is the calculation for the whole matrix: Step 2: Matrix of Cofactors This is easy! Just apply a "checkerboard" of minuses to the "Matrix of Minors". In other words, you need to change the sign of alternate cells, like this:

23 BBA 120 Business Mathematics Step 3: Adjugate (also called Adjoint) Now "Transpose" all elements of the previous matrix... in other words swap their positions over the diagonal (the diagonal stays the same): Step 4: Multiply by 1/Determinant Now find the determinant of the original matrix. This isn't too hard, because we already calculated the determinants of the smaller parts when we did "Matrix of Minors". The adjugate of a matrix A is the transpose of the cofactor matrix C of A:. So: multiply the top row elements by their matching "minor" determinants: Determinant = = 10 And now multiply the Adjugate by 1/Determinant: And we are done! Summary For a 3 3 matrix the matrix inverse is 21

24 22 Unit 6: Matrix Algebra POINTS TO REMEMBER For each element, calculate the determinant of the values not on the row or column, to make the Matrix of Minors Apply a checkerboard of minuses to make the Matrix of Cofactors Transpose to make the Adjugate Multiply by 1/Determinant to make the Inverse Larger Matrices It is exactly the same steps for larger matrices (such as a 4 4, 5 5, etc), but wow! there is a lot of calculation involved. For a 4 4 Matrix you have to calculate determinants. So it is often easier to use computers (such as the Matrix Calculator.) More Examples For a 2 2 generic matrix, The adjugate of the 2 2 matrix is It is seen that det(adj(a)) = det(a) and adj(adj(a)) = A.. For a 3 3 generic matrix Consider the matrix

25 BBA 120 Business Mathematics Its adjugate is the transpose of the cofactor matrix So that we have where. Therefore C, the matrix of cofactors for A, is The adjugate is the transpose of the cofactor matrix. Thus, for instance, the (3,2) entry of the adjugate is the (2,3) cofactor of A. (In this example, C happens to be its own transpose, so adj(a) = C.) 23

26 24 Unit 6: Matrix Algebra Cramer s Rule Cramer's Rule Recall the general 3 4 matrix used to solve systems of three equations: This matrix will be used to solve systems by Cramer's Rule. We divide it into four separate 3 3 matrices: This rule for solving equations is known as Cramer s rule, named after the Swiss mathematician Gabriel Cramer ( ), although it was initially invented by Leibniz. It may be easily generalized to the case of solving n linear equations in n unknowns. Cramer showed promise in mathematics from an early age. At 18 he received his doctorate and at 20 he was cochair of mathematics at the University of Geneva. a 1 b 1 c 1 D = [ a 2 b 2 c 2 ] a 3 b 3 c 3 d 1 b 1 c 1 D x = [ d 2 b 2 c 2 ] d 3 b 3 c 3 a 1 d 1 c 1 D y = [ a 2 d 2 c 2 ] a 3 d 3 c 3 a 1 b 1 d 1 D z = [ a 2 b 2 d 2 ] a 3 b 3 d 3 D is the 3 3 coefficient matrix, and D x, D y, and D z are each the result of substituting the constant column for one of the coefficient columns in D. Note: If detd = 0 and D x, detd y, or detd z 0, the system is inconsistent. If detd = 0 and D x = detd y = detd z = 0, the system has multiple solutions. Cramer's Rule states that: x = detd x detd y = detd y detd z = detd z detd Thus, to solve a system of three equations with three variables using Cramer's Rule,

27 BBA 120 Business Mathematics 1. Arrange the system in the following form: 2. Create D, D x, D y, and D z. a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 3. Find detd, detd x, detd y, and detd z 4. x = detd x detd, y = detd y detd, and z = detd z detd. Example: Solve the following system: Solution: 1. Rearrange the system: 2. Create the matrices: 8x + 10z = 7y x + 3y + 8z = 7 5y + 9 = 4x + 2z 8x 7y + 10z = 15 2x + 3y + 8z = 7 4x + 5y 2z = D = [ ] D x = [ ] D y = [ ] D z = [ ] Find the determinants: detd = ( ) ( ) = = 48 detd x = ( ) ( ) = =

28 26 Unit 6: Matrix Algebra detd y = ( ) ( ) = 772 ( 916) = 144 detd z = ( ) ( ) = = x = = 7, y = = 3, z = = 2 Thus, (x, y, z) = (7, 3, 2) Exercises: 1. Can you use Cramer s rule to solve a linear system with a 3 2 coefficient matrix? Explain. 2. Can you use Cramer s rule to solve a linear system with a 4 4 coefficient matrix? Explain. 3. Evaluate each second-order determinant i ii ii iii Evaluate each third-order determinant i ii Use Crammers Rule to solve the following systems of equations. i. ii. iii. 2x 3y + z = 3 4x + 3y + 2z = 11 x y z = 3 x + 4y 3z = 25 3x + y z = 2 4x + y + 2z = 1 12x 14y + 11z = 5 15x + 7y 9z = 13 5x 3y + 2z = 0 6. REVENUE ANALYSIS A company manufactures ten-speed and threespeed bicycles. The weekly demand equations are

29 BBA 120 Business Mathematics p = x + 5y q = x 4y where $p is the price of a ten-speed bicycle, $q is the price of a threespeed bicycle, x is the weekly demand for ten-speed bicycles, and y is the weekly demand for three-speed bicycles. The weekly revenue R is given by R = xp + yp (A) Use system (above) to express the daily revenue in terms of x and y only. (B) Use Cramer s rule to solve system for x and y in terms of p and q, and then express the daily revenue R in terms of p and q only. 27