778 CHAPTER 11 Systems of Equations and Inequalities The matrix method is especially effective for systems of equations for which the number of equations and the number of variables are unequal. Here, too, such a system is either inconsistent or consistent. If it is consistent, it will have either exactly one solution or infinitely many solutions. Solving a System of Linear Equations Using Matrices Solve: x y z x y z y z x y z 2 0 2 2 3 3 1 4 2 13 − + = + − =− − =− − + + = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ ⎪⎪ (1) (2) (3) (4) Solution EXAMPLE 9 This is a system of four equations containing three variables. Begin with the augmented matrix, and use row operations to write the matrix in row echelon form. 1 2 1 2 2 3 0 1 1 1 4 2 0 3 1 13 1 2 1 0 6 5 0 1 1 0 2 3 0 3 1 13 1 2 1 0 1 1 0 6 5 0 2 3 0 1 3 13 − − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → − − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ↑ ↑ =− + = + R r r R r r 2 2 1 2 4 1 4 Interchange rows 2 and 3. 1 2 1 0 1 1 0 0 1 0 0 5 0 1 3 15 1 2 1 0 1 1 0 0 1 0 0 0 0 1 3 0 → − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ↑ ↑ =− + =− + R r r R r r 6 2 3 2 3 4 2 4 =− + R r r 5 4 3 4 Since the matrix is in row echelon form, we could now back-substitute = z 3 to find x and y. Or we can continue and obtain the reduced row echelon form. 1 2 1 0 1 1 0 0 1 0 0 0 0 1 3 0 1 0 1 0 1 1 0 0 1 0 0 0 2 1 3 0 1 0 0 0 1 0 0 0 1 0 0 0 1 2 3 0 − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ → ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ↑ ↑ = + R r r 2 1 2 1 = + = + R r r R r r 1 3 1 2 3 2 The matrix is now in reduced row echelon form, and we can see that the solution is = = = x y z 1, 2, 3 or, using an ordered triplet, ( ) 1, 2, 3 . Solution The bottom row is equivalent to the equation + + = − x y z 0 0 0 27 which has no solution. The original system is inconsistent. Begin with the augmented matrix, and use row operations to write the matrix in row echelon form. 1 1 1 2 1 1 1 2 2 6 3 0 1 1 1 0 3 3 0 1 1 6 9 6 1 1 1 0 1 1 0 3 3 6 6 9 1 1 1 0 1 1 0 0 0 6 6 27 − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ↑ ↑ ↑ =− + =− + R r r R r r 2 1 2 1 2 3 1 3 Interchange rows 2 and 3. = + R r r 3 3 2 3 Now Work PROBLEM 29 Now Work PROBLEM 71
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