897 9.2 Matrix Solution of Linear Systems The solution set, written with z arbitrary, is 5116z + 38, 7z + 15, z26. S Now Try Exercise 43. x - 16z = 38 x = 16z + 38 Add 16z. y - 7z = 15 y = 7z + 15 Add 7z. Summary of Possible Cases When matrix methods are used to solve a system of linear equations and the resulting matrix is written in diagonal form (or as close as possible to diagonal form), there are three possible cases. 1. If the number of rows with nonzero elements to the left of the vertical line is equal to the number of variables in the system, then the system has a single solution. See Examples 1 and 2. 2. If one of the rows has the form 30 0 g 0 a4 with a≠0, then the system has no solution. See Example 3. 3. If there are fewer rows in the matrix containing nonzero elements than the number of variables, then the system has either no solution or infinitely many solutions. If there are infinitely many solutions, give the solutions in terms of one or more arbitrary variables. See Example 4. EXAMPLE 4 Solving a System with Infinitely Many Solutions Use the Gauss-Jordan method to solve the system. Write the solution set with z arbitrary. 2x - 5y + 3z = 1 x - 2y - 2z = 8 SOLUTION Recall from the previous section that a system with two equations in three variables usually has an infinite number of solutions. We can use the Gauss-Jordan method to give the solution with z arbitrary. c 2 1 -5 -2 3 -2 2 1 8d Write the augmented matrix. c 1 2 -2 -5 -2 3 2 8 1d Interchange rows to obtain 1 in the first row, first column position. c 1 0 -2 -1 -2 7 2 8 -15d -2R1 + R2 c 1 0 -2 1 -2 -7 2 8 15d -1R2 c 1 0 0 1 -16 -7 2 38 15d 2R2 + R1 It is not possible to go further with the Gauss-Jordan method. The equations that correspond to the final matrix are x - 16z = 38 and y - 7z = 15. Solve these equations for x and y, respectively. 1 7 0 1 1 0
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