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INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2007 Pearson Education Asia Chapter 6 Matrix Algebra

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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

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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

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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 Objectives

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2007 Pearson Education Asia Matrices Matrix Addition and Scalar Multiplication Matrix Multiplication Solving Systems by Reducing Matrices (continued) Inverses Leontiefs InputOutput Analysis 6.1) 6.2) 6.3) 6.4) Chapter 6: Matrix Algebra Chapter Outline 6.5) 6.6) 6.7)

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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. For the entry a ij, we call i the row subscript and j the column subscript.

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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

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.1 Matrices Example 3 – Constructing Matrices Equality of Matrices Matrices A = [a ij ] and B = [b ij ] are equal if they have the same size and a ij = b ij for each i and j. Transpose of a Matrix A transpose matrix is denoted by A T. If, find. Solution: Observe that.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 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.

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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:

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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

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.2 Matrix Addition and Scalar Multiplication Example 5 – Matrix Subtraction a. b.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 1 – Sizes of Matrices and Their Product AB is the m× p matrix C whose entry c ij 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.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 3 – Matrix Products a. b. c. d.

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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

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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.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.3 Matrix Multiplication Example 9 – Raw Materials and Cost Find QRC when Solution:

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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:

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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

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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

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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.

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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

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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

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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 x 4 takes on any real value.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (continued) Example 1 – Two-Parameter Family of Solutions Using matrix reduction, solve Solution: The matrix is reduced to The solution is

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.5 Solving Systems by Reducing Matrices (Continue) The system is called a homogeneous system if c 1 = c 2 = … = c m = 0. The system is non-homogeneous if at least one of the cs is not equal to 0. Concept for number of solutions: 1.k < n infinite solutions 2.k = n unique solution

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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.

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.6 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.

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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.

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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:

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.7 Leontiefs 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.)

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2007 Pearson Education Asia Chapter 6: Matrix Algebra 6.7 Leontiefs 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

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