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7.3 MULTIVARIABLE LINEAR SYSTEMS Copyright © Cengage Learning. All rights reserved. What You Should Learn • Use back-substitution to solve linear systems in row-echelon form. • Use Gaussian elimination to solve systems of linear equations. • Solve nonsquare systems of linear equations. • Use systems of linear equations in three or more variables to model and solve real-life problems. 2 Row-Echelon Form and Back-Substitution 3 Row-Echelon Form and Back-Substitution System of Three Linear Equations in Three Variables: x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17 Equivalent System in Row-Echelon Form: x – 2y + 3z = 9 y + 3z = 5 z=2 4 Example 1 – Using Back-Substitution in Row-Echelon Form Solve the system of linear equations. x – 2y + 3z = 9 Equation 1 y + 3z = 5 Equation 2 z=2 Equation 3 Solution: y + 3(2) = 5 y = –1. Substitute 2 for z. Solve for y. 5 Example 1 – Solution x – 2(–1) + 3(2) = 9 x = 1. cont’d Substitute –1 for y and 2 for z. Solve for x. The solution is x = 1, y = –1, and z = 2, which can be written as the ordered triple (1, –1, 2). 6 Gaussian Elimination 7 Gaussian Elimination 8 Example 3 – Using Gaussian Elimination to Solve a System Solve the system of linear equations. x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17 Equation 1 Equation 2 Equation 3 Solution: Add Equation 1 to Equation 2. 9 Example 3 – Solution cont’d x – 2y + 3z = 9 y + 3z = 5 2x – 5y + 5z = 17 Multiply Equation 1 by –2. Write Equation 3. Add revised Equation 1 to Equation 3. 10 Example 3 – Solution cont’d x – 2y + 3z = 9 y + 3z = 5 –y – z = –1 x – 2y + 3z = 9 y + 3z = 5 2z = 4 x – 2y + 3z = 9 y + 3z = 5 z=2 x = 1, y = –1, and z = 2. 11 Gaussian Elimination 12 Gaussian Elimination Solution: one point Solution: one line Solution: one plane Solution: none Solution: none 13 Nonsquare Systems 14 Nonsquare Systems Square System: # equations = # unknowns Nonsquare System: the number of equations differs from the number of variables. A system of linear equations cannot have a unique solution unless there are at least as many equations as there are variables in the system. 15 Example 6 – A System with Fewer Equations than Variables Solve the system of linear equations. x – 2y + z = 2 2x – y – z = 1 Equation 1 Equation 2 Solution: x – 2y + z = 2 3y – 3z = –3 x – 2y + z = 2 y – z = –1 16 Example 6 – Solution cont’d y = z – 1. x – 2y + z = 2 x – 2(z – 1) + z = 2 x – 2z + 2 + z = 2 x=z 17 Example 6 – Solution cont’d Let z = a, where a is a real number, you have the solution x = a, y = a – 1, and z = a. So, every ordered triple of the form (a, a – 1, a) a is a real number. is a solution of the system. 18 Example 2x 4 y z 7 2 x 4 y 2 z 6 x 4y z 0 19 Applications 20 Example 7 – Vertical Motion The height at time t of an object that is moving in a (vertical) line with constant acceleration a is given by the position equation The height s is measured in feet, the acceleration a is measured in feet per second squared, t is measured in seconds, v0 is the initial velocity (at t = 0), and s0 is the initial height. 21 Example 7 – Vertical Motion cont’d Find the values of a, v0, and s0 if s = 52 at t = 1, s = 52 at t = 2, and s = 20 at t = 3, and interpret the result. (See Figure 7.17.) Figure 7.17 22 Example 7 – Solution By substituting the three values of t and s into the position equation, you can obtain three linear equations in a, v0, and s0. When t = 1: a(1)2 + v0(1) + s0 = 52 When t = 2: a(2)2 + v0(2) + s0 = 52 2a + 2v0 + s0 = 52 When t = 3: a(3)2 + v0(3) + s0 = 20 9a + 6v0 + 2s0 = 40 a + 2v0 + 2s0 = 104 This produces the following system of linear equations. a + 2v0 + 2s0 = 104 2a + 2v0 + s0 = 52 9a + 6v0 + 2s0 = 40 23 Example 7 – Solution cont’d Now solve the system using Gaussian elimination. a + 2v0 + 2s0 = 104 – 2v0 – 3s0 = –156 9a + 6v0 + 2s0 = 40 a + 2v0 + 2s0 = 104 – 2v0 – 3s0 = –156 – 12v0 – 16s0 = –896 a + 2v0 + 2s0 = 104 – 2v0 – 3s0 = –156 2s0 = 40 24 Example 7 – Solution cont’d a + 2v0 + 2s0 = 104 v0 + s0 = 78 s0 = 20 So, the solution of this system is a = –32, v0 = 48, and s0 = 20, which can be written as (–32, 48, 20). This solution results in a position equation of s = –16t2 + 48t + 20 and implies that the object was thrown upward at a velocity of 48 feet per second from a height of 20 feet. 25 Group Assignment x 2 y 2 Dx Ey F 0 Find the equation of the circle that passes through the points (-3,-1), (2,4), (-6, 8) 26