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Unformatted text preview: MAT 202 Spring 2004, Quiz 1 on Chapters 1 and 2 1. (8 pts) Find all solutions to the given system by using elementary row operations to bring the matrix of the system into reduced row echelon form. Be sure to indicate which variables are free variables and which are leading variables. x + 2 y + u + v + 3 w = 2 x + 4 y + 5 u + 2 v + 3 w = 3 x + 6 y + 3 u 4 v + 2 w = Since the system is homogeneous (only zero appears on the right hand side) we can take the coefficient matrix and reduce it completely to solve. 1 2 1 1 3 2 4 5 2 3 3 6 3 3 9 → 1 2 3 1 1 1 1 We conclude that x = 2 y 3 w , u = w and v = w . The variables x , u and v are leading variables since they correspond to lead 1’s in the reduced matrix. The other variables y and w are free. In vector form the solution is all vectors in R 5 of the form x y u v w = s  2 1 + t  3 1 1 1 2. (10 pts) Find all vectors ~ b so that the system below is solvable. Interpret your answer geo metrically. 2 x 1 x 2 + 4 x 3 = b 1 x 1 + x 2 x 3 = b 2 7 x 1 + x 2 + 5 x 3 = b 3 Form the augmented matrix and reduce: 2 1 4  b 1 1 1 1  b 2 7 1 5  b 3 →  3 6  b 1 2 b 2 1 1 1  b 2 6 12  b 3 7 b 2 → 1 1 1  b 2 1 2  2 b 2 / 3 b 1 / 3 1 2  7 b 2 / 6 b 3...
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This note was uploaded on 04/28/2008 for the course MAT 202 taught by Professor Staff during the Spring '08 term at Princeton.
 Spring '08
 Staff
 Linear Algebra, Algebra

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