HW 1 08

HW 1 08 - MA 511 Prof. Jaroslaw Wlodarczyk Spring 2008...

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MA 511 Spring 2008 Prof. Jaroslaw Wlodarczyk Khalil Yousef MA 511 - Khalil Yousef Homework 1 Problem Set 1.2: 4) 3 by 2 system 1 2 2 2 = = = + y y x y x Solution: As we can see the system is not resolvable because not all the lines intersect together in a common point. If the set of equations rights-hand sides are zero, as follows: 0 0 0 2 = = = + y y x y x , then there is no any non-zero choice of right hand sides that allows the three lines to intersect at the same point because all the three lines intersects only at the origin (0,0).
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MA 511 Spring 2008 Prof. Jaroslaw Wlodarczyk Khalil Yousef MA 511 - Khalil Yousef 2 Problem Set 1.3: 2) 11 9 10 1 3 2 = + = + y x y x : Problem 1 Equations Solution: Step 1 (Elimination): subtract 5 times equation 1 from equation 2: 6 6 1 3 2 = = + y y x 1 = y , So using back-substitution 2 2 ) 1 ( 3 1 2 3 1 = = = y x Verification: ok y x = = = + 11 1 9 20 3 4 9 3 1 10 2 2 11 1 ? 9 3 10 2 If the right-hand side changes to (4, 44), then the new solution is? 44 9 10 4 3 2 = + = + y x y x Elimination: Eq 2 =Eq 2 – 5Eq 1 24 6 4 3 2 = = + y y x 4 = y 8 2 ) 4 ( 3 4 2 3 4 = = = y x 4) g dy cx f by ax = + = + Solution: a c Problem Set 1.4: 2) 3 1 1 6 1 5 1 4 , 0 1 0 9 8 7 6 5 4 3 2 1 , 3 1 2 1 9 8 6 6 3 4
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MA 511 Spring 2008 Prof. Jaroslaw Wlodarczyk Khalil Yousef MA 511 - Khalil Yousef 3 Solution: = + + + = + = 9 8 7 3 6 3 5 3 4 1 1 1 3 6 5 4 1 3 1 1 6 1 5 1 4 = + + + + + + = + + = 8 5 2 0 8 0 0 5 0 0 2 0 9 6 3 0 8 5 2 1 7 4 1 0 0 1 0 9 8 7 6 5 4 3 2 1 = + + + = + = 7 5 3 3 4 2 3 1 2 9 6 3 3 1 8 6 4 2 1 9 8 6 6 3 4 3 1 2 1 4) Solution: First case: 1 1 . × × × = m n n m Ax x A 1 × m Separate Multiplication. Second case: p m p n n m AB B A × × × = . p m × Separate Multiplication. 22) Write down the 3 by 3 matrices that produce these elimination steps.
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HW 1 08 - MA 511 Prof. Jaroslaw Wlodarczyk Spring 2008...

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