774 CHAPTER 11 Systems of Equations and Inequalities In Words To obtain an augmented matrix in row echelon form: • Add rows, exchange rows, or multiply a row by a nonzero constant. • Work from top to bottom and left to right. • Get 1’s in the main diagonal with 0’s below the 1’s. • Once the entry in row 1, column 1 is 1 with 0’s below it, do not use row 1 in your row operations. • Once the entries in row 1, column 1 and row 2, column 2 are 1 with 0’s below, do not use row 1 or 2 in your row operations (and so on). Matrix Method for Solving a System of Linear Equations (Row Echelon Form) Step 1 Write the augmented matrix that represents the system. Step 2 Use row operations to obtain 1 in row 1, column 1. Step 3 Use row operations that leave row 1 unchanged, but change the entries in column 1 below row 1 to 0’s. Step 4 Use row operations to obtain 1 in row 2, column 2, but leave the entries in columns to the left unchanged. If it is impossible to place 1 in row 2, column 2, place 1 in row 2, column 3. Once the 1 is in place, use row operations to obtain 0’s below it. (Place any rows that contain only 0’s on the left side of the vertical bar, at the bottom of the matrix.) Step 5 Now repeat Step 4 to obtain 1 in the next row, but one column to the right. Continue until the bottom row or the vertical bar is reached. Step 6 The matrix that results is the row echelon form of the augmented matrix. Analyze the system of equations corresponding to it to solve the original system. Step 3 Use row operations that leave row 1 unchanged, but change the entries in column 1 below row 1 to 0’s. That is, we want a =0 21 and a = .0 31 Step 4 Use row operations to obtain 1 in row 2, column 2, and 0’s below it. That is, we want a =1 22 and a = .0 32 Step 5 Repeat Step 4 to obtain 1 in row 3, column 3. That is, we want a = .1 33 Step 6 The matrix on the right in Step 5 is the row echelon form of the augmented matrix. Use back-substitution to solve the original system. Next, we want 0 in row 2, column 1 and 0 in row 3, column 1. Use the row operations = − + R r r 2 2 1 2 and = − + R r r 3 . 3 1 3 Note that row 1 is unchanged using these row operations. − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ 1 2 3 1 2 4 1 0 1 1 6 13 1 0 0 1 0 1 1 2 4 1 4 10 ↑ =− + R r r 2 2 1 2 =− + R r r 3 3 1 3 The third row of the augmented matrix represents the equation = − z 2. Using = − z 2, back-substitute into the equation − = y z4 10 (row 2) and obtain y z y y 4 10 4 2 10 2 ( ) − = − − = = =− z 2 Solve for y. Finally, back-substitute = y 2 and = − z 2 into the equation + + = x y z 1 (row 1) and obtain ( ) + + = + + − = = x y z x x 1 2 2 1 1 = =− y z 2, 2 Solve for x. The solution of the system is = = = − x y z 1, 2, 2 or, using an ordered triplet, ( ) − 1, 2, 2 . We want the entry in row 2, column 2 to be 1. We also want to have 0 below the 1 in row 2, column 2. Interchanging rows 2 and 3 will accomplish both goals. − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ 1 0 0 1 0 1 1 2 4 1 4 10 1 0 0 1 1 0 1 4 2 1 10 4 To obtain 1 in row 3, column 3, use the row operation = − R r 1 2 . 3 3 The result is − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ 1 0 0 1 1 0 1 4 2 1 10 4 1 0 0 1 1 0 1 4 1 1 10 2 ↑ =− R r 1 2 3 3
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