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Chapter 9 Matrices and Determinants © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved 1 SECTION 9.1 Matrices and Systems of Equations OBJECTIVES 1 2 3 4 Define a matrix. Use matrices to solve a system. Use Gaussian elimination to solve a system. Use Gauss–Jordan elimination to solve a system. © 2010 Pearson Education, Inc. All rights reserved 2 DEFINITION OF A MATRIX A matrix is a rectangular array of numbers denoted by a11 a 21 A a m1 a12 a22 am 2 ... a1n ... a2 n ... amn Row 1 Row 2 Row m Column 1 Column 2 Column n © 2010 Pearson Education, Inc. All rights reserved 3 DEFINITION OF A MATRIX If a matrix A has m rows and n columns, then A is said to be of order m by n (written m × n). The entry or element in the ith row and jth column is a real number and is denoted by the double-subscript notation aij. We can call aij the (i, j)th entry. © 2010 Pearson Education, Inc. All rights reserved 4 DEFINITION OF A MATRIX If A has n rows and n columns, then A is called a square matrix of order n. The entries a11, a22, …, ann form the main diagonal of A. A 1 × n matrix is called a row matrix, and an n × 1 matrix is called a column matrix. © 2010 Pearson Education, Inc. All rights reserved 5 EXAMPLE 1 Determining the Order of Matrices Determine the order of each matrix. Identify square, row, and column matrices. Identify entries in the main diagonal of each square matrix. © 2010 Pearson Education, Inc. All rights reserved 6 EXAMPLE 1 Determining the Order of Matrices Solution a. Matrix A with one row and one column is a 1 × 1 matrix. A is a square matrix of order 1. In A, a11 = 3. A is also a column and a row matrix. b. Matrix B with one row and three columns is a 1 × 3 matrix. B is a row matrix. © 2010 Pearson Education, Inc. All rights reserved 7 EXAMPLE 1 Determining the Order of Matrices Solution c. Matrix C, a 2 × 2 matrix, is a square matrix of order 2. In C, the entries c11 = 0 and c22 = 4 form the main diagonal. d. Matrix D, a 3 × 3 matrix, is a square matrix of order 3. In D, the entries d11 = 1, d22 = 5, and d33 = 9 form the main diagonal. © 2010 Pearson Education, Inc. All rights reserved 8 MATRIX AND LINEAR SYSTEMS We can display the constants and coefficients of a system in matrix called the augmented matrix of the system. Constants x y z 1 2x 3y z 10 x y 2z 0 1 1 1 1 2 3 1 1 0 1 1 2 0 Coefficients of x Coefficients of y Coefficients of z © 2010 Pearson Education, Inc. All rights reserved 9 ELEMENTARY ROW OPERATIONS Two matrices are row equivalent if one can be obtained from the other by a sequence of elementary row operations. Row Operation In Symbols Interchange two rows Ri n Rj Description Interchange the ith and jth rows. Multiply a row by a cRj Multiply the jth row by c, nonzero constant c ≠ 0. Add a multiple of one cRi + Rj g Rj Replace the jth row by row to another row adding c times ith row to it. © 2010 Pearson Education, Inc. All rights reserved 10 EXAMPLE 3 Applying Elementary Row Operations Perform the indicated row operations (a), (b), and (c) in order on the following matrix: Solution a. Interchange the 1st and 2nd rows. © 2010 Pearson Education, Inc. All rights reserved 11 EXAMPLE 3 Applying Elementary Row Operations Solution continued b. Multiply the 1st row by . c. Add –3 times the 1st row to the 2nd row. © 2010 Pearson Education, Inc. All rights reserved 12 ROW-ECHELON FORM AND REDUCED ROW-ECHELON FORM An m × n matrix is in row-echelon form if it has the following three properties: 1. The leading entry of each nonzero row is 1. 2. The leading entry in a row is to the right of the leading entry in the row above it. 3. All nonzero rows are above the rows consisting entirely of zeros. © 2010 Pearson Education, Inc. All rights reserved 13 ROW-ECHELON FORM AND REDUCED ROW-ECHELON FORM A matrix in row-echelon form having the following property is in reduced row-echelon form: Each leading 1 is the only nonzero entry in its column. © 2010 Pearson Education, Inc. All rights reserved 14 SOLVING LINEAR SYSTEMS BY USING GAUSSIAN ELIMINATION Step 1 Write the augmented matrix. Step 2 Use elementary row operations to transform the augmented matrix into row-echelon form. Step 3 Write the system of linear equations that corresponds to the last in Step 2. Step 4 Use the system of equations from Step 3, together with back-substitution, to find the system’s solution set. © 2010 Pearson Education, Inc. All rights reserved 15 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solve by Gaussian elimination. 2x y z 6 3x 4 y 2 z 4 x y z 2 Solution Step 1 The augmented matrix of the system is. 2 1 1 6 A 3 4 2 4 1 1 1 2 © 2010 Pearson Education, Inc. All rights reserved 16 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solution continued Step 2 1 R1 R3 3 2 1 1 2 4 2 4 1 1 6 1 1 1 2 3R1 R2 R2 0 1 1 2 2 R1 R3 R3 0 1 3 10 © 2010 Pearson Education, Inc. All rights reserved 17 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solution continued Step 2 continued 1 1 1 2 1 R2 0 1 1 2 0 1 3 10 1 1 1 2 R2 R3 R3 0 1 1 2 0 0 4 12 © 2010 Pearson Education, Inc. All rights reserved 18 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solution continued Step 2 continued 1 1 1 2 The matrix is now 1 in row-echelon R3 0 1 1 2 4 form. 0 0 1 3 Step 3 The system corresponding to the last matrix in Step 2 is x y z 2 (1) y z 2 (2) z 3 (3) © 2010 Pearson Education, Inc. All rights reserved 19 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solution continued Step 4 Equation (3) in Step 3 gives the value z = 3. Back-substitute z = 3 in equation (2). yz2 y3 2 y 1 Back-substitute z = 3 and y = –1 in equation (1). x y z 2 x 1 3 2 x2 © 2010 Pearson Education, Inc. All rights reserved 20 EXAMPLE 6 Solving a System by Using Gaussian Elimination Solution continued The solution set for the system is {(2, –1, 3)}. You should check the solution by substituting these values for x, y, and z into the original system of equations. © 2010 Pearson Education, Inc. All rights reserved 21 EXAMPLE 7 Attempting to Solve a System with No Solution Solve the system of equations by Gaussian elimination. Solution Step 1 The augmented matrix of the system is. © 2010 Pearson Education, Inc. All rights reserved 22 EXAMPLE 7 Attempting to Solve a System with No Solution Solution continued Step 2 © 2010 Pearson Education, Inc. All rights reserved 23 EXAMPLE 7 Attempting to Solve a System with No Solution Solution continued Step 2 continued The matrix is now in row-echelon form. © 2010 Pearson Education, Inc. All rights reserved 24 EXAMPLE 7 Attempting to Solve a System with No Solution Solution continued Step 3 The system corresponding to the last matrix in Step 2 is A false statement Because the equation 0 = 1 is never true, we conclude that this system is inconsistent. Step 4 The solution set for the system is . © 2010 Pearson Education, Inc. All rights reserved 25 EXAMPLE 8 Solving a System of Equations by GaussJordan Elimination Solve the system given in Example 4 by GaussJordan elimination. x y z 1 2 x 3 y z 10 The given system x y 2z 0 Solution The augmented matrix of the system is © 2010 Pearson Education, Inc. All rights reserved 26 EXAMPLE 8 Solving a System of Equations by GaussJordan Elimination Solution continued row-echelon form © 2010 Pearson Education, Inc. All rights reserved 27 EXAMPLE 8 Solving a System of Equations by GaussJordan Elimination Solution continued reduced row-echelon form The corresponding system of equations for the last augmented matrix is The solution set is {(5, 1, 3)}. © 2010 Pearson Education, Inc. All rights reserved 28 EXAMPLE 9 Solving a System with Infinitely Many Solutions Solve the system of equations. Solution The augmented matrix of the system is © 2010 Pearson Education, Inc. All rights reserved 29 EXAMPLE 9 Solving a System with Infinitely Many Solutions Solution continued We need a zero at the (3, 1) position. And we need a zero at the (1, 2) position. reduced row-echelon form © 2010 Pearson Education, Inc. All rights reserved 30 EXAMPLE 9 Solving a System with Infinitely Many Solutions Solution continued The equivalent system is Solving for x and y in terms of z, we obtain Each real number z results in a solution with y = –4z + 4 and x = 3z – 4, giving infinitely many solutions of the form (3z – 4, –4z + 4, z). The solution set is {(3z – 4, –4z + 4, z)}. © 2010 Pearson Education, Inc. All rights reserved 31 EXAMPLE 10 Leontief Input–Output Model Consider an economy that has steel, coal, and transportation industries. There are two types of demands (measured in dollars) on the production of each industry: interindustry demand and external consumer demand. The outputs and requirements of the three industries are shown in on the next slide. For example, $1.00 of transportation output requires $0.10 from steel and $0.01 from coal. © 2010 Pearson Education, Inc. All rights reserved 32 EXAMPLE 10 Leontief Input–Output Model © 2010 Pearson Education, Inc. All rights reserved 33 EXAMPLE 10 Leontief Input–Output Model a. Write a system of equations that expresses the outputs of the three industries. Assume that all quantities are given in millions of dollars. b. Verify that s = 15, c = 10, and t = 8 will meet both interindustry and consumer demand. © 2010 Pearson Education, Inc. All rights reserved 34 EXAMPLE 10 Leontief Input–Output Model Solution a. To satisfy both consumer and interindustry demand, we obtain the following system of outputs. (s denotes total steel output; c, total coal output; t, total transportation output.) © 2010 Pearson Education, Inc. All rights reserved 35 EXAMPLE 10 Leontief Input–Output Model Solution continued You can rewrite this system as: b. To verify that s = 15, c = 10, and t = 8 satisfy those equations, we substitute them into each equation. © 2010 Pearson Education, Inc. All rights reserved 36