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INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences Chapter 6 Matrix Algebra 2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics 2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus 2007 Pearson Education Asia Chapter 6: Matrix Algebra Chapter Objectives • Concept of a matrix. • Special types of matrices. • Matrix addition and scalar multiplication operations. • Express a system as a single matrix equation using matrix multiplication. • Matrix reduction to solve a linear system. • Theory of homogeneous systems. • Inverse matrix. • Use a matrix to analyze the production of sectors of an economy. 2007 Pearson Education Asia Chapter 6: Matrix Algebra Chapter Outline 6.1) Matrices 6.2) Matrix Addition and Scalar Multiplication 6.3) Matrix Multiplication 6.4) Solving Systems by Reducing Matrices 6.5) Solving Systems by Reducing Matrices (continued) 6.6) Inverses 6.7) Leontief’s Input—Output Analysis 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices • A matrix consisting of m horizontal rows and n vertical columns is called an m×n matrix or a matrix of size m×n. a11 a12 a 21 a12 . . . . . . am1 am 2 ... a1n ... a2n ... . ... . ... . ... amn • For the entry aij, we call i the row subscript and j the column subscript. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices Example 1 – Size of a Matrix a. The matrix 1 2 0has size 1 3 . 1 6 b. The matrix 5 1 has size 3 2 . 9 4 c. The matrix 7 has size 1 1 . 7 2 4 1 3 d. The matrix 9 11 5 6 8 has size3 5 . 6 2 1 1 1 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices Equality of Matrices • Matrices A = [aij ] and B = [bij] are equal if they have the same size and aij = bij for each i and j. Transpose of a Matrix • A transpose matrix is denoted by AT. Example 3 – Constructing Matrices If A 1 2 3 , find AT . 4 5 6 Solution: 1 4 AT 2 5 3 6 Observe that AT T 2007 Pearson Education Asia A. Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Matrix Addition • Sum A + B is the m × n matrix obtained by adding corresponding entries of A and B. Example 1 – Matrix Addition a. 1 4 7 2 1 7 2 2 8 0 2 5 6 4 3 6 4 4 3 8 3 6 3 0 5 3 6 0 8 6 b. 1 2 2 3 4 1 is impossible as matrices are not of the same size. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Example 3 – Demand Vectors for an Economy Demand for the consumers is D1 3 2 5 D2 0 17 1 D3 4 6 12 For the industries is DC 0 1 4 DE 20 0 8 DS 30 5 0 What is the total demand for consumers and the industries? Solution: D1 D2 D3 3 2 5 0 17 1 4 6 12 7 25 18 DC DE DS 0 1 4 20 0 8 30 5 0 50 6 12 Total: 7 25 18 50 6 12 57 31 30 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Scalar Multiplication • Properties of Scalar Multiplication: Subtraction of Matrices • Property of subtraction is A 1A 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Example 5 – Matrix Subtraction 6 6 2 2 6 6 2 4 8 4 1 4 1 4 4 1 1 8 0 3 2 0 3 3 0 2 3 3 1 a. 2 6 2 6 6 0 8 0 1 2 4 2 5 b. AT 2B 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication • AB is the m× p matrix C whose entry cij is given by n c ij aik bkj ai 1b1 j ai 2b2 j ... ain bnj k 1 Example 1 – Sizes of Matrices and Their Product A = 3 × 5 matrix B = 5 × 3 matrix AB = 3 × 3 matrix but BA = 5 × 5 matrix. C = 3 × 5 matrix D = 7 × 3 matrix CD = undefined but DC = 7 × 5 matrix. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 3 – Matrix Products a. 4 1 2 35 32 6 b. 1 1 6 21 6 2 12 3 3 18 c. 0 2 16 3 11 1 3 0 1 2 2 1 2 1 3 10 1 0 1 0 4 2 1 2 7 4 10 d. a11 a12 b11 b12 a11b11 a12 b21 a11b12 a12 b22 a b a b a b a b a b a b 22 21 21 12 22 22 21 22 21 22 21 11 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 5 – Cost Vector Given the price and the quantities, calculate the total cost. 7 units of A P 2 3 4 Q 5 units of B 11 units of C Solution: The cost vector is 7 PQ 2 3 4 5 73 11 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 7 – Associative Property If 1 2 A 3 4 1 0 C 0 2 1 1 3 0 1 B 1 1 2 compute ABC in two ways. Solution 1: 1 0 1 2 3 0 1 ABC 1 1 2 0 2 3 4 1 1 1 2 2 1 4 9 3 4 6 19 3 4 Solution 2: 1 0 1 2 3 0 1 0 2 AB C 3 4 1 1 2 1 1 1 0 1 2 5 4 9 0 2 6 19 5 4 11 1 1 Note that A(BC) = (AB)C. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 9 – Raw Materials and Cost Find QRC when Q 5 7 9 5 20 16 7 17 R 7 18 12 9 21 6 25 8 5 13 Solution: 2500 1200 C 800 150 1500 2500 5 20 16 7 17 1200 75850 RC 7 18 12 9 21 800 81550 6 25 8 5 13 150 71650 1500 75850 QRC QRC 5 7 1281550 1,809,900 71650 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 11 – Matrix Operations Involving I and O If 3 2 A 1 4 52 B 1 10 51 3 10 1 0 I 0 1 0 0 O 0 0 compute each of the following. Solution: 1 0 3 2 2 2 a. I A 0 1 1 4 1 3 3 2 3 2 2 0 3 6 1 0 b. 3A 2I 3 2 3 3 6 1 4 0 1 1 4 0 2 3 2 0 0 c. AO O 1 4 0 0 3 2 52 d. AB 1 1 4 10 51 1 0 I 3 0 1 10 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 13 – Matrix Form of a System Using Matrix Multiplication 2x 5 x2 4 Write the system 1 8 x1 3 x2 7 in matrix form by using matrix multiplication. Solution: 2 5 If A 8 3 x1 X x2 4 B 7 then the single matrix equation is AX B 2 5 x1 4 8 3 x 7 2 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Elementary Row Operations 1. Interchanging two rows of a matrix 2. Multiplying a row of a matrix by a nonzero number 3. Adding a multiple of one row of a matrix to a different row of that matrix 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Properties of a Reduced Matrix • All zero-rows at the bottom. • For each nonzero-row, leading entry is 1 and the rest zeros. • Leading entry in each row is to the right of the leading entry in any row above it. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Example 1 – Reduced Matrices For each of the following matrices, determine whether it is reduced or not reduced. 1 0 a. 0 3 0 0 0 d. 0 0 0 1 0 0 b. 0 1 0 1 0 0 e.0 0 0 0 1 0 Solution: a. Not reduced c. Not reduced e. Not reduced 2007 Pearson Education Asia 0 1 c. 1 0 0 1 0 3 f.0 0 1 2 0 0 0 0 b. Reduced d. Reduced f. Reduced Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Example 3 – Solving a System by Reduction By using matrix reduction, solve the system 2x 3y 1 2x y 5 x y 1 Solution: Reducing the augmented coefficient matrix of the system, 2 3 1 2 1 5 1 1 1 x4 We have y 3 2007 Pearson Education Asia 1 0 4 0 1 3 0 0 0 Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Example 5 – Parametric Form of a Solution Using matrix reduction, solve 2x1 3 x2 2x3 6 x 4 0 x 2 2 x3 x 4 2 3 x1 3 x3 6 x 4 9 Solution: Reducing the matrix of the system, 2 3 2 6 10 0 1 2 1 2 3 0 3 6 9 We have x1 4 52 x 4 x2 0 x 1 1 x 2 4 3 2007 Pearson Education Asia 1 0 0 52 4 0 1 0 0 0 0 0 1 21 1 and x4 takes on any real value. Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (continued) Example 1 – Two-Parameter Family of Solutions Using matrix reduction, solve x1 2x2 5 x3 5 x 4 3 x1 x2 3 x3 4 x 4 1 x x x 2x 3 2 3 4 1 Solution: 1 The matrix is reduced to 0 0 1 3 1 1 2 1 2 0 0 0 0 0 The solution is 2007 Pearson Education Asia x1 1 r 3s x 2 2r s 2 x3 r x 4 s Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (Continue) • • The system a11x1 a12 x 2 ... a1n x n c1 . . . . am1x1 am 2 x 2 ... amn xn c m is called a homogeneous system if c1 = c2 = … = cm = 0. The system is non-homogeneous if at least one of the c’s is not equal to 0. Concept for number of solutions: 1. k < n infinite solutions 2. k = n unique solution 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (Continue) Example 3 – Number of Solutions of a Homogeneous System Determine whether the system has a unique solution or infinitely many solutions. x y 2z 0 2x 2y 4z 0 Solution: 2 equations (k), homogeneous system, 3 unknowns (n). The system has infinitely many solutions. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.6 Inverses • When matrix CA = I, C is an inverse of A and A is invertible. Example 1 – Inverse of a Matrix Let 1 2 A 3 7 2 3 1 and C 7 . Determine whether C is an inverse of A. 7 2 1 2 1 0 Solution: CA I 3 1 3 7 0 1 Thus, matrix C is an inverse of A. 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.6 Inverses Method to Find the Inverse of a Matrix • When matrix is reduced, A I R B , - If R = I, A is invertible and A−1 = B. - If R I, A is not invertible. Example 3 – Determining the Invertibility of a Matrix 1 0 A 2 2 Determine if Solution: We have is invertible. 1 0 1 0 A I 2 2 0 1 Matrix A is invertible where 2007 Pearson Education Asia 1 0 1 0 I B 1 0 1 1 2 1 0 A 1 1 2 1 Chapter 6: Matrix Algebra 6.6 Inverses Example 5 – Using the Inverse to Solve a System Solve the system by finding the inverse of the coefficient matrix. x 2x 1 1 3 4 x1 2x2 x3 2 x 2x 10 x 1 2 3 1 Solution: 1 We have A 4 0 2 2 1 1 2 10 For 9 2 2 inverse, A1 412 4 92 5 1 1 The solution is given by X = 2007 Pearson Education Asia x1 9 2 2 1 7 A−1B:x2 412 4 92 2 17 x3 5 1 1 1 4 Chapter 6: Matrix Algebra 6.7 Leontief’s Input-Output Analysis • • Entries are called input–output coefficients. Use matrices to show inputs and outputs. Example 1 – Input-Output Analysis Given the input–output matrix, suppose final demand changes to be 77 for A, 154 for B, and 231 for C. Find the output matrix for the economy. (The entries are in millions of dollars.) 2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.7 Leontief’s Input-Output Analysis Example 1 – Input-Output Analysis Solution: Divide entries by the total value of output to get A: Final-demand matrix: Output matrix is 77 D 154 231 692.5 1 X I A D 380 495 2007 Pearson Education Asia