776 CHAPTER 11 Systems of Equations and Inequalities Solving a Dependent System of Linear Equations Using Matrices Solve: x y z x y z x y z 6 4 12 2 2 8 5 3 − − = − + + =− + − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) Solution EXAMPLE 7 Start with the augmented matrix of the system and use row operations to obtain 1 in row 1, column 1 with 0’s below. 6 1 1 12 2 2 5 1 1 4 8 3 1 2 0 12 2 2 5 1 1 1 8 3 1 2 0 0 22 2 0 11 1 1 4 2 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ ↑ =− + R r r 1 1 3 1 Using = z 1, back-substitute to get x y y x y y 1 8 12 1 15 7 3 → − + = − ⋅ =− ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ − = =− ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ (1) (2) (1) (2) Simplify. Using = − y 3, back-substitute into − = x y 7 to get = x 4. The solution of the system is = = − = x y z 4, 3, 1 or, using an ordered triplet, ( ) − 4, 3, 1 . Using = − y 3 from the second equation, back-substitute into − = x y 2 3 6 to get = x 4. The solution of the system is x y 4, 3, = = − z 1 = or, using an ordered triplet, ( ) − 4, 3, 1 . Notice that the row echelon form of the augmented matrix in the graphing solution differs from the row echelon form in the algebraic solution, yet both matrices provide the same solution! This is because the two solutions used different row operations to obtain the row echelon form. In all likelihood, the two solutions parted ways in Step 4 of the algebraic solution, where we avoided introducing fractions by interchanging rows 2 and 3. Sometimes it is advantageous to write a matrix in reduced row echelon form. In this form, row operations are used to obtain entries that are 0 above (as well as below) the leading 1 in a row. For example, the row echelon form obtained in the algebraic solution to Example 6 is 1 0 0 1 1 0 1 12 1 8 15 1 − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ To write this matrix in reduced row echelon form, proceed as follows: 1 0 0 1 1 0 1 12 1 8 15 1 1 0 0 0 1 0 11 12 1 7 15 1 1 0 0 0 1 0 0 0 1 4 3 1 − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ ↑ = + R r r 1 2 1 The matrix is now written in reduced row echelon form. The advantage of writing the matrix in this form is that the solution to the system, = = − = x y z 4, 3, 1, is readily found, without the need to back-substitute. Another advantage will be seen in Section 11.4, where the inverse of a matrix is discussed. The method used to write a matrix in reduced row echelon form is called Gauss-Jordan elimination. Most graphing utilities also have the ability to put a matrix in reduced row echelon form. Figure 9(a) shows the reduced row echelon form of the augmented matrix from Example 6 using the RREF command on a TI-84 Plus CE graphing calculator. Figure 9(b) shows the same result using the Desmos Matrix Calculator. Figure 9 Reduced row echelon form (b) (a) = + = + R r r R r r 11 12 1 3 1 2 3 2 Now Work PROBLEMS 39 AND 49 The matrix method for solving a system of linear equations also identifies systems that have infinitely many solutions and systems that are inconsistent. = + =− + R r r R r r 12 5 2 1 2 3 1 3
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