Example find the solution set of the following system

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Elementary and Intermediate Algebra: Algebra Within Reach
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Chapter 8 / Exercise 41
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
Example Find the solution set of the following system of linear equations: . 0 8 2 6 = + = + + = + + z y x z y x z y x Rb Ra cRa Ra Solution Since the system of linear equations is already in standard form we can write down its augmented matrix immediately. We then perform a sequence of elementary row operations until a row- equivalent matrix is obtained from which we can readily determine the solution for the given system of equations. The following notation will be used to indicate the row operation(s) being performed on the matrices at each stage of the process: means interchange rows a and b. means replace row a by c × row a. 8
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Elementary and Intermediate Algebra: Algebra Within Reach
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Chapter 8 / Exercise 41
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
cRb Ra Ra + means replace row a by the sum of row a and c × row b. R2 - R1 R1 R1 - R3 R3 R1 - R2 R2 6 - 2 - 0 0 2 0 1 0 6 1 1 1 0 1 - 1 1 2 0 1 0 6 1 1 1 0 1 1 1 8 1 2 1 6 1 1 1 1 0 0 1 4 1 0 1 4 1 0 1 3 1 0 0 2 0 1 0 3 1 0 0 2 0 1 0 6 2 0 0 2 0 1 0 R3 - R1 R1 R3 2 -1 R3 The system of equations that corresponds to this last matrix is 3 2 1 = = = z y x . Therefore, the solution to the original system of linear equations is the ordered triple (1, 2, 3). Example Use elementary row operations on the augmented matrix to solve the following system of linear equations: 9 4 3 2 6 3 2 = + + = + + z y x z y x 3 = + + z y x 0 1 0 1 3 1 1 1 3 1 1 1 R2 R3 R3 R2 R1 R1 2R1 R3 R3 R1 R2 R2 Solution 0 0 0 0 3 2 1 0 3 2 1 0 3 2 1 0 9 4 3 2 6 3 2 1 0 (1) 2 3 (2) x z y z . The equations corresponding to this last matrix are 0 0 (3) = + = = . = These equations may be rewritten as 0 0 = 2 3 = z y z x . We see that x and y depend on z and can write the solution as ( z , 3 – 2 z , z ) where z is arbitrary. This system has infinitely many solutions, each value of z produces a new solution. Example Use elementary row operations on the augmented matrix to solve the following system of linear equations. 10 4 3 2 6 3 2 3 = + + = + + + + z y x z y x z y x = Vector Geometry and Linear Algebra MATH 1300 Unit 3 9
Solution . 1 0 0 0 3 2 1 0 3 1 1 1 4 2 1 0 3 2 1 0 3 1 1 1 10 4 3 2 6 3 2 1 3 1 1 1 R2 R3 R3 2R1 R3 R3 R1 R2 R2 3 0 2 3 0 0 0 1 x y z x y z The equations corresponding to this last matrix are x y z + + = + + = + + = . Consider the last of these three equations. Clearly there are no values for x , y and z that will satisfy this last equation and so the system has no solution. 3.4 Problems 1. Use elementary row operations on the augmented matrices to find the solution set for each of the following systems of linear equations. a. b. 1 = y x 5 = + y x 1 3 2 2 = + + y x = y x 2 c. 4 2 2 = y x = y x 1 = y x 2 6 = + = z y x 6 5 3 3 = + + d. 12 3 2 = + y x e. f. 2 0 = + z y x + + z y x 7 6 4 3 = + + = + + z y x z y x 2 3 6 1 z x y z x y + z y x 3 = + + z y x + = g. 2 2 x y z + = + = h. x 2 = + z x 1 = + z y 3.5 The Gauss-Jordan elimination procedure In this section a systematic procedure for solving systems of linear equations is introduced. This procedure is called the Gauss-Jordan elimination procedure. The system of linear equations is

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