SECTION 11.2 Systems of Linear Equations: Matrices 775 Solving a System of Linear Equations Using Matrices (Row Echelon Form) Solve: x y z x y z x y z 8 2 3 2 3 2 9 9 − + = + − =− − − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) EXAMPLE 6 Algebraic Solution Step 1 The augmented matrix of the system is 1 2 3 1 3 2 1 1 9 8 2 9 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ Step 2 Because the entry 1 is already present in row 1, column 1, go to Step 3. Step 3 Perform the row operations = − + R r r 2 2 1 2 and = − + R r r 3 . 3 1 3 Each of these leaves the entry 1 in row 1, column 1 unchanged, while causing 0’s to appear under it. 1 2 3 1 3 2 1 1 9 8 2 9 1 0 0 1 5 1 1 3 12 8 18 15 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ =− + R r r 2 2 1 2 =− + R r r 3 3 1 3 Step 4 The easiest way to obtain the entry 1 in row 2, column 2 without altering column 1 is to interchange rows 2 and 3 (another way would be to multiply row 2 by 1 5 , but this introduces fractions). 1 0 0 1 1 5 1 12 3 8 15 18 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ To get a 0 under the 1 in row 2, column 2, perform the row operation = − + R r r 5 . 3 2 3 1 0 0 1 1 5 1 12 3 8 15 18 1 0 0 1 1 0 1 12 57 8 15 57 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ =− + R r r 5 3 2 3 Step 5 Continuing, obtain a 1 in row 3, column 3 by using = R r 1 57 . 3 3 1 0 0 1 1 0 1 12 57 8 15 57 1 0 0 1 1 0 1 12 1 8 15 1 − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ = R r 1 57 3 3 Step 6 The matrix on the right is the row echelon form of the augmented matrix. The system of equations represented by the matrix in row echelon form is x y z y z z 8 12 15 1 − + = − =− = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) Graphing Solution The augmented matrix of the system is 1 2 3 1 3 2 1 1 9 8 2 9 − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ Enter this matrix into a graphing utility and name it A. See Figure 8(a). Using the REF (Row Echelon Form) command on matrix A gives the results shown in Figure 8(b). The system of equations represented by the matrix in row echelon form is x y z y z z 2 3 3 3 15 13 24 13 1 − − = + =− = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎪⎪⎪ ⎩ ⎪⎪ ⎪⎪ ⎪⎪⎪ (1) (2) (3) Using = z 1, back-substitute to get (1) (2) − − ⋅ = + ⋅ =− ⎧ ⎨ ⎪⎪ ⎪⎪⎪ ⎩ ⎪⎪ ⎪⎪ ⎪ x y y 2 3 3 1 3 15 13 1 24 13 (1) (2) − = = − = − ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y y 2 3 6 39 13 2 (continued) (continued) Figure 8 Row echelon form (b) (b) (a)
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