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Chapter6 matrixalgebra-151003151111-lva1-app6892
- 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 6Chapter 6 Matrix AlgebraMatrix Algebra
- 2. ©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
- 3. ©2007 Pearson Education Asia 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 INTRODUCTORY MATHEMATICAL ANALYSIS
- 4. ©2007 Pearson Education Asia • 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. Chapter 6: Matrix Algebra Chapter ObjectivesChapter Objectives
- 5. ©2007 Pearson Education Asia Matrices Matrix Addition and Scalar Multiplication Matrix Multiplication Solving Systems by Reducing Matrices Solving Systems by Reducing Matrices (continued) Inverses Leontief’s Input—Output Analysis 6.1) 6.2) 6.3) 6.4) Chapter 6: Matrix Algebra Chapter OutlineChapter Outline 6.5) 6.6) 6.7)
- 6. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices6.1 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. • For the entry aij, we call i the row subscript and j the column subscript. mnmm n n aaa aaa aaa ... ...... ...... ...... ... ... 21 21221 11211
- 7. ©2007 Pearson Education Asia a. The matrix has size . b. The matrix has size . c. The matrix has size . d. The matrix has size . Chapter 6: Matrix Algebra 6.1 Matrices Example 1 – Size of a Matrix [ ]021 31× − 49 15 61 23× [ ]7 11× −− − 11126 865119 42731 53×
- 8. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices Example 3 – Constructing 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 . If , find . Solution: Observe that . = 654 321 A = 63 52 41 T A ( ) AA TT = T A
- 9. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication6.2 Matrix Addition and Scalar Multiplication Example 1 – Matrix Addition Matrix Addition • Sum A + B is the m × n matrix obtained by adding corresponding entries of A and B. a. b. is impossible as matrices are not of the same size. −= ++ +− −+ = − − + 68 83 08 0635 4463 2271 03 46 27 63 52 41 + 1 2 43 21
- 10. ©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 For the industries is What is the total demand for consumers and the industries? Solution: Total: [ ] [ ] [ ]12641170523 321 === DDD [ ] [ ] [ ]05308020410 === SEC DDD [ ] [ ] [ ] [ ]1825712641170523321 =++=++ DDD [ ] [ ] [ ] [ ]1265005308020410 =++=++ SEC DDD [ ] [ ] [ ]3031571265018257 =+
- 11. ©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 ( )AA 1−=−
- 12. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Example 5 – Matrix Subtraction a. b. − − − = −+ −−− +− = − − − 13 08 84 3203 1144 2662 30 14 26 23 14 62 −− = − − − =− 52 80 42 66 10 26 2BAT
- 13. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication6.3 Matrix Multiplication Example 1 – Sizes of Matrices and Their Product • AB is the m× p matrix C whose entry cij is given by 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. njinji n k jikjikij babababac +++== ∑= ...22 1 11
- 14. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 3 – Matrix Products a. b. c. d. [ ] [ ]32 6 5 4 321 = [ ] = 183 122 61 61 3 2 1 −− − − = − −− − − 1047 0110 11316 212 312 201 401 122 031 ++ ++ = 2222122121221121 2212121121121111 2221 1211 2221 1211 babababa babababa bb bb aa aa
- 15. ©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. Solution: The cost vector is [ ]432=P Cofunits Bofunits Aofunits 11 5 7 =Q [ ] [ ]73 11 5 7 432 = =PQ
- 16. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 7 – Associative Property If compute ABC in two ways. Solution 1: Solution 2: Note that A(BC) = (AB)C. = − = −− − = 11 20 01 211 103 43 21 CBA ( ) −− = − −− − = −− − = 196 94 43 12 43 21 11 20 01 211 103 43 21 BCA ( ) −− = − −− = −− − = 196 94 11 20 01 1145 521 11 20 01 211 103 43 21 CAB
- 17. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 9 – Raw Materials and Cost Find QRC when Solution: [ ]975=Q = 1358256 21912187 17716205 R = 1500 150 800 1200 2500 C = = 71650 81550 75850 1500 150 800 1200 2500 1358256 21912187 17716205 RC ( ) [ ] [ ]900,809,1 71650 81550 75850 1275 = == RCQQRC
- 18. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 11 – Matrix Operations Involving I and O If compute each of the following. Solution: 00 00 10 01 41 23 10 3 10 1 5 1 5 2 = = − − = = OIBA 31 22 41 23 10 01 a. −− −− = − =− AI ( ) = − = − =− 63 63 20 02 41 23 3 10 01 2 41 23 323b. IA OAO = = 00 00 41 23 c. IAB = = − − = 10 01 41 23 d. 10 3 10 1 5 1 5 2
- 19. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 13 – Matrix Form of a System Using Matrix Multiplication Write the system in matrix form by using matrix multiplication. Solution: If then the single matrix equation is =+ =+ 738 452 21 21 xx xx = = = 7 4 38 52 2 1 B x x XA = = 7 4 38 52 2 1 x x BAX
- 20. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices6.4 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
- 21. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices6.4 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.
- 22. ©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. Solution: a. Not reduced b. Reduced c. Not reduced d. Reduced e. Not reduced f. Reduced 0000 2100 3010 f. 010 000 001 e. 000 000 d. 01 10 c. 010 001 b. 30 01 a.
- 23. ©2007 Pearson Education Asia 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 Solution: Reducing the augmented coefficient matrix of the system, We have − 1 5 1 11 12 32 =+ =+ −=+ 1 52 132 yx yx yx − 0 3 4 00 10 01 −= = 3 4 y x
- 24. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.4 Solving Systems by Reducing Matrices Example 5 – Parametric Form of a Solution Using matrix reduction, solve Solution: Reducing the matrix of the system, We have and x4 takes on any real value. − 9 2 10 6303 1210 6232 =+− =++ =+++ 9633 22 06232 431 432 4321 xxx xxx xxxx 1 0 4 100 0010 001 2 1 2 5 −= = −= 42 1 3 2 42 5 1 1 0 4 xx x xx
- 25. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices6.5 Solving Systems by Reducing Matrices (continued)(continued) Example 1 – Two-Parameter Family of Solutions Using matrix reduction, solve Solution: The matrix is reduced to The solution is =+−− −=+++ −=+++ 32 143 3552 4321 4321 4321 xxxx xxxx xxxx − 0 2 1 0000 1210 3101 = = −−−= −−= sx rx srx srx 4 3 2 1 22 31
- 26. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (Continue) • The system 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. =+++ =+++ mnmnmm nn cxaxaxa cxaxaxa ... . . . . ... 2211 11212111 Concept for number of solutions: 1. k < n infinite solutions 2. k = n unique solution
- 27. ©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. Solution: 2 equations (k), homogeneous system, 3 unknowns (n). The system has infinitely many solutions. =−+ =−+ 0422 02 zyx zyx
- 28. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.6 Inverses6.6 Inverses Example 1 – Inverse of a Matrix • When matrix CA = I, C is an inverse of A and A is invertible. Let and . Determine whether C is an inverse of A. Solution: Thus, matrix C is an inverse of A. = 73 21 A − − = 13 27 C ICA = = − − = 10 01 73 21 13 27
- 29. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.6 Inverses Example 3 – Determining the Invertibility of a Matrix Determine if is invertible. Solution: We have Matrix A is invertible where Method to Find the Inverse of a Matrix • When matrix is reduced, , - If R = I, A is invertible and A−1 = B. - If R ≠ I, A is not invertible. [ ] [ ]BRIA →→ = 22 01 A [ ] = 10 01 22 01 IA [ ]BI= − 2 1 1 01 10 01 − =− 2 1 1 1 01 A
- 30. ©2007 Pearson Education Asia 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. Solution: We have For inverse, The solution is given by X = A−1 B: −=−+ =+− =− 1102 224 12 321 321 31 xxx xxx xx − − − = 1021 124 201 A − − =− 115 4 229 2 9 2 411 A − − − = − − − − = 4 17 7 1 2 1 115 4 229 2 9 2 41 3 2 1 x x x
- 31. ©2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.7 Leontief’s Input-Output Analysis6.7 Leontief’s Input-Output Analysis Example 1 – Input-Output Analysis • Entries are called input–output coefficients. • Use matrices to show inputs and outputs. 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.)
- 32. ©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 = 231 154 77 D ( ) =−= − 495 380 5.692 1 DAIX