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wksht09sol

wksht09sol - MATH 152 L1 L2 Applied Linear Algebra and...

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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 = 0 , 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 0 3 - 6 7 - 1 2 - 4 3 1 - 3 R 1 + R 2 ------→ - 2 R 1 + R 3 1 - 2 2 0 0 0 1 - 1 0 0 - 1 1 - 2 R 2 + R 1 ------→ R 2 + R 3 1 - 2 0 2 0 0 1 - 1 0 0 0 0 . 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 , 0 = 0 . 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 = 0 . Solution We simplify the augmented matrix as follows 0 0 - 1 - 3 2 2 - 2 4 - 5 7 1 1 1 - 2 3 - 2 1 0 2 R 3 + R 2 -----→ R 1 R 3 1 - 2 3 - 2 1 0 0 0 1 3 3 1 0 0 - 1 - 3 2 2 - 3 R 2 + R 1 ------→ R 2 + R 3 1 - 2 0 - 11 - 8 - 3 0 0 1 3 3 1 0 0 0 0 5 3 1 5 R 3 ----→ 1 - 2 0 - 11 - 8 - 3 0 0 1 3 3 1 0 0 0 0 1 3 / 5 8 R 3 + R 1 ------→ - 3 R 3 + R 2 1 - 2 0 - 11 0 9 / 5 0 0 1 3 0 - 4 / 5 0 0 0 0 1 3 / 5 . Interpreted the last matrix (in reduced row echelon form) as a system of linear equations, we have x 1 - 2 x 2 - 11 x 4 = 9 / 5 , x 3 +3 x 4 = - 4 / 5 , x 5 = 3 / 5 .

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