SECTION 11.2 Systems of Linear Equations: Matrices 777 Now Work PROBLEM 55 We can obtain 1 in row 2, column 2 without altering column 1 by using either = − R r 1 22 2 2 or = + R r r 23 11 . 2 3 2 We use the first of these here. 1 2 0 0 22 2 0 11 1 1 4 2 1 2 0 0 1 1 11 0 11 1 1 2 11 2 1 2 0 0 1 1 11 0 0 0 1 2 11 0 − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ↑ ↑ =− R r 1 22 2 2 =− + R r r 11 3 2 3 This matrix is in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of only two equations. x y y z 2 1 1 11 2 11 − = − =− ⎧ ⎨ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪ (1) (2) To make it easier to write some of the solutions, we express both x and y in terms of z. From the second equation, = − y z 1 11 2 11 . Now back-substitute this solution for y into the first equation to get ( ) = + = − + = + x y z z 2 1 2 1 11 2 11 1 2 11 7 11 The original system is equivalent to the system x z y z z 2 11 7 11 1 11 2 11 where canbeanyrealnumber = + = − ⎧ ⎨ ⎪⎪ ⎪⎪⎪ ⎩ ⎪⎪ ⎪⎪ ⎪ (1) (2) Let’s look at the situation.The original system of three equations is equivalent to a system containing two equations.This means that any values of x y z , , that satisfy both = + = − x z y z 2 11 7 11 and 1 11 2 11 are solutions. For example, = = = − = = = − z x y z x y 0, 7 11 , 2 11 ; 1, 9 11 , 1 11 ; and = − = = − z x y 1, 5 11 , 3 11 are three of the solutions of the original system. There are, in fact, infinitely many values of x y , , and z for which the two equations are satisfied. That is, the original system has infinitely many solutions. We write the solution of the original system as x z y z z 2 11 7 11 1 11 2 11 where canbeanyrealnumber = + = − ⎧ ⎨ ⎪⎪ ⎪⎪⎪ ⎩ ⎪⎪ ⎪⎪ ⎪ or, using ordered triplets, as x y z x z y z z , , 2 11 7 11 , 1 11 2 11 , anyrealnumber { } ( ) = + = − Identifying an Inconsistent System of Linear Equations Using Matrices Solve: + + = − − = + + = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y z x y z x y z 6 2 3 2 2 0 EXAMPLE 8 (continued)
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