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**Unformatted text preview: **MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 2004-05 ¥ Week 09 Worksheet Solutions : Systems of Linear Equations Q1. (Method of reduction) Solve the system of linear equations x 1- 2 x 2 +2 x 3 = , 3 x 1- 6 x 2 +7 x 3 =- 1 , 2 x 1- 4 x 2 +3 x 3 = 1 . Solution We simplify the augmented matrix as follows 1- 2 2 3- 6 7- 1 2- 4 3 1 - 3 R 1 + R 2------→- 2 R 1 + R 3 1- 2 2 1- 1- 1 1 - 2 R 2 + R 1------→ R 2 + R 3 1- 2 2 1- 1 . Interpreted the last matrix (in reduced row echelon form) as a system of linear equations, we have x 1- 2 x 2 = 2 , x 3 =- 1 , = . The last equation is redundant. From the second equation, we have x 3 =- 1. From the first equation, we have x 1 = 2 + 2 x 2 . The system has non-unique solutions ( x 1 ,x 2 ,x 3 ) = (2 + 2 r, r,- 1) , for any arbitrary number r . Q2. (Method of reduction) Solve the system of linear equations - x 3- 3 x 4 +2 x 5 = 2 ,- 2 x 1 +4 x 2- 5 x 3 +7 x 4 + x 5 = 1 , x 1- 2 x 2 +3 x 3- 2 x 4 + x 5 = . Solution We simplify the augmented matrix as follows - 1- 3 2 2- 2 4- 5 7 1 1 1- 2 3- 2 1 2 R 3 + R 2-----→ R 1 ↔ R 3 1- 2 3- 2 1 1 3 3 1- 1- 3 2 2 - 3 R 2 + R 1------→ R 2 + R 3 1- 2- 11- 8- 3 1 3 3 1 5 3 1 5 R 3----→ 1- 2- 11- 8- 3 1 3 3 1 1 3 / 5 8 R 3 + R 1------→- 3 R 3 + R 2 1- 2- 11 9 / 5 1 3- 4 / 5 1 3 / 5 ....

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