Linear Equations in Linear Algebra PDF

1 Linear Equations in Linear Algebra 1.1 SYSTEMS OF LINEAR EQUATIONS © 2016 Pearson Education, Inc. LINEAR EQUATION ▪ A linear equation in the variables x1 , , xn is an equation that can be written in the form a1 x1 + a2 x2 +...

1 Linear Equations in Linear Algebra 1.1 SYSTEMS OF LINEAR EQUATIONS © 2016 Pearson Education, Inc. LINEAR EQUATION ▪ A linear equation in the variables x1 , , xn is an equation that can be written in the form a1 x1 + a2 x2 + + an xn = b where b and the coefficients a1 , , an are real or complex numbers, usually known in advance. ▪ A system of linear equations (or a linear system) is a collection of one or more linear equations involving the same variables — say, x1,…., xn. © 2016 Pearson Education, Inc. Slide 1.1- 2 LINEAR EQUATION ▪ A solution of the system is a list (s1, s2,…, sn) of numbers that makes each equation a true statement when the values s1,…, sn are substituted for x1,…, xn, respectively. ▪ The set of all possible solutions is called the solution set of the linear system. ▪ Two linear systems are called equivalent if they have the same solution set. © 2016 Pearson Education, Inc. Slide 1.1- 3 LINEAR EQUATION ▪ A system of linear equations has 1. no solution, or 2. exactly one solution, or 3. infinitely many solutions. ▪ A system of linear equations is said to be consistent if it has either one solution or infinitely many solutions. ▪ A system is inconsistent if it has no solution. © 2016 Pearson Education, Inc. Slide 1.1- 4  MATRIX NOTATION ▪ The essential information of a linear system can be recorded compactly in a rectangular array called a matrix. Given the system, x1 − 2 x2 + x3 = 0 2 x2 − 8 x3 = 8 −4 x1 + 5 x2 + 9 x3 = −9, with the coefficients of each variable aligned in columns, the matrix  1 −2 1  0 2 −8    −4 5 9  Is called the coefficient matrix (or matrix of coefficients) of the system © 2016 Pearson Education, Inc. Slide 1.1- 5 MATRIX NOTATION ▪ An augmented matrix of a system consists of the coefficient matrix with an added column containing the constants from the right sides of the equations. ▪ For the given system of equations,  1 −2 1 0  0 2 −8 8    −4 5 9 −9  is called the augmented matrix of the system. © 2016 Pearson Education, Inc. Slide 1.1- 6 MATRIX SIZE ▪ The size of a matrix tells how many rows and columns it has. If m and n are positive integers, an m n matrix is a rectangular array of numbers with m rows and n columns. (The number of rows always comes first.) ▪ The basic strategy for solving a linear system is to replace one system with an equivalent system (i.e., one with the same solution set) that is easier to solve. © 2016 Pearson Education, Inc. Slide 1.1- 7 SOLVING SYSTEM OF EQUATIONS ▪ Example 1: Solve the given system of equations. x1 − 2 x2 + x3 = 0 ----(1) 2 x2 − 8 x3 = 8 ----(2) −4 x + 5 x + 9 x = −9 ----(3) 1 2 3 ▪ Solution: The elimination procedure is shown here with and without matrix notation, and the results are placed side by side for comparison. © 2016 Pearson Education, Inc. Slide 1.1- 8 SOLVING SYSTEM OF EQUATIONS x1 − 2 x2 + x3 = 0  1 −2 1 0  0 2 −8 8 2 x2 − 8 x3 = 8   −4 x1 + 5 x2 + 9 x3 = −9  −4 5 9 −9  ▪ Keep x1 in the first equation and eliminate it from the other equations. To do so, add 4 times equation 1 to equation 3. 4 x − 8 x + 4 x = 0 1 2 3 −4 x1 + 5 x2 + 9 x3 = −9 © 2016 Pearson Education, Inc. −3 x2 + 13 x3 = −9 Slide 1.1- 9 SOLVING SYSTEM OF EQUATIONS ▪ The result of this calculation is written in place of the original third equation: x1 − 2 x2 + x3 = 0  1 −2 1 0 0 2 −8 8 2 x2 − 8 x3 = 8   −3x2 + 13x3 = −9 0 −3 13 −9  ▪ Now, multiply equation 2 by 1/ 2 in order to obtain 1 as the coefficient for x2. © 2016 Pearson Education, Inc. Slide 1.1- 10 SOLVING SYSTEM OF EQUATIONS x1 − 2 x2 + x3 = 0  1 −2 1 0 x2 − 4 x3 = 4 0 1 −4 4    −3x2 + 13x3 = −9 0 −3 13 −9  ▪ Use the x2 in equation 2 to eliminate the −3x2 in equation 3. 3x − 12 x = 12 2 3 −3x2 + 13x3 = −9 x3 = 3 © 2016 Pearson Education, Inc. Slide 1.1- 11 SOLVING SYSTEM OF EQUATIONS ▪ The new system has a triangular form. x1 − 2 x2 + x3 = 0  1 −2 1 0 x2 − 4 x3 = 4 0 1 −4 4    x3 = 3 0 0 1 3 ▪ Eventually, you want to eliminate the −2x2 term from equation 1, but it is more efficient to use the x3 term in equation 3 first to eliminate the −4x3 and x3 terms in equations 2 and 1. © 2016 Pearson Education, Inc. Slide 1.1- 12 SOLVING SYSTEM OF EQUATIONS 4 x3 = 12 − x3 = −3 x2 − 4 x3 = 4 x1 − 2 x2 + x3 = 0 x2 = 16 x1 − 2 x2 = −3 ▪ Now, combine the results of these two operations. x1 − 2 x2 = −3  1 −2 0 −3 0 1 0 16  x2 = 16   x3 = 3 0 0 1 3 © 2016 Pearson Education, Inc. Slide 1.1- 13 SOLVING SYSTEM OF EQUATIONS ▪ Move back to the x2 in equation 2, and use it to eliminate the −2x2 above it. Because of the previous work with x3, there is now no arithmetic involving x3 terms. Add 2 times equation 2 to equation 1 and obtain the system: x1 = 29  1 0 0 29  x2 = 16 0 1 0 16    x3 = 3 0 0 1 3 © 2016 Pearson Education, Inc. Slide 1.1- 14 SOLVING SYSTEM OF EQUATIONS ▪ Thus, the only solution of the original system is (29,16,3). To verify that (29,16,3) is a solution, substitute these values into the left side of the original system, and compute. (29) − 2(16) + (3) = 29 − 32 + 3 = 0 2(16) − 8(3) = 32 − 24 = 8 −4(29) + 5(16) + 9(3) = −116 + 80 + 27 = −9 ▪ The results agree with the right side of the original system, so (29,16,3) is a solution of the system. © 2016 Pearson Education, Inc. Slide 1.1- 15 ELEMENTARY ROW OPERATIONS ▪ Elementary row operations include the following: 1. (Replacement) Replace one row by the sum of itself and a multiple of another row. 2. (Interchange) Interchange two rows. 3. (Scaling) Multiply all entries in a row by a nonzero constant. ▪ Two matrices are called row equivalent if there is a sequence of elementary row operations that transforms one matrix into the other. © 2016 Pearson Education, Inc. Slide 1.1- 16 ELEMENTARY ROW OPERATIONS ▪ It is important to note that row operations are reversible. ▪ If the augmented matrices of two linear systems are row equivalent, then the two systems have the same solution set. ▪ Two fundamental questions about a linear system are as follows: 1. Is the system consistent; that is, does at least one solution exist? 2. If a solution exists, is it the only one; that is, is the solution unique? © 2016 Pearson Education, Inc. Slide 1.1- 17 EXISTENCE AND UNIQUENESS OF SYSTEM OF EQUATIONS ▪ Example 3: Determine if the following system is consistent: x2 − 4 x3 = 8 2 x1 − 3x2 + 2 x3 = 1 (5) 5 x1 − 8 x2 + 7 x3 = 1 ▪ Solution: The augmented matrix is 0 1 −4 8  2 −3 2 1    5 −8 7 1 © 2016 Pearson Education, Inc. Slide 1.1- 18 EXISTENCE AND UNIQUENESS OF SYSTEM OF EQUATIONS ▪ To obtain an x1 in in the first equation, interchange rows 1 and 2: 2 −3 2 1   0 1 −4 8    5 −8 7 1 ▪ To eliminate the 5x1 term in the third equation, add −5 / 2 times row 1 to row 3. 2 −3 2 1 0 1 −4 8 (6)    0 −1/ 2 2 −3 / 2  © 2016 Pearson Education, Inc. Slide 1.1- 19 EXISTENCE AND UNIQUENESS OF SYSTEM OF EQUATIONS ▪ Next, use the x2 term in the second equation to eliminate the −(1/ 2)x2 term from the third equation. Add 1/ 2times row 2 to row 3.  2 −3 2 1 0 1 −4 8 (7)    0 0 0 5 / 2  ▪ The augmented matrix is now in triangular form. To interpret it correctly, go back to equation notation. 2 x1 − 3x2 + 2 x3 = 1 x2 − 4 x3 = 8 (8) © 2016 Pearson Education, Inc. 0 =5/ 2 Slide 1.1- 20 EXISTENCE AND UNIQUENESS OF SYSTEM OF EQUATIONS ▪ The equation 0 = 5 / 2 is a short form of 0 x1 + 0 x2 + 0 x3 = 5 / 2. ▪ There are no values of x1, x2, x3 that satisfy (8) because the equation 0 = 5 / 2 is never true. ▪ Since (8) and (5) have the same solution set, the original system is inconsistent (i.e., has no solution). © 2016 Pearson Education, Inc. Slide 1.1- 21