Sp11_Ex2_Part_1_soln - EGN 3420 Exam 2 Part 1 solution Sp...

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Unformatted text preview: EGN 3420 Exam 2, Part 1 solution Sp 2011 Do Any 7 problems! (No calculators) Write your answers on the Answer Sheet only after you are sure of your answer. Make sure you have only 7 answers on the Answer Sheet! All problems are worth 10 pts 1. The system of equations x y 0 x y 1 a) consistent with a single solution Ax b , 1 1 A , 1 1 is b) consistent with infinite solutions 0 b , 1 x x y c) inconsistent 11 1 1 2 | A | 1 1 Since | A | is not zero, the equations are consistent with a unique solution. x 2. The system of equations x 2y 0 is 1 2x a) consistent with a single solution Ax b , 1 1 A 1 2 , 2 1 y y 1 b) consistent with infinite solutions 1 b 0 1 1 1 1 1 1 1 1 1 1 1 2 0 0 1 1 0 1 1 (A | b ) 2 1 1 0 3 1 0 0 4 From the last row of the echelon form, the equations are inconsistent. x 3. The system of equations y z 2 x x 2y y z 2z 0 is K c) inconsistent a) inconsistent for all values of K b) consistent for all values of K c) consistent with solution x 1, y 0, z 1 when K 1 d) none of the above Ax b , 1 1 1 A 1 2 1 , 1 1 2 2 b 0 , K x x y z x 1 1 1 2 111 2 1 1 2 1 0 0 1 2 0 ( A | b) 2 1 1 2 K 0 2 1 K 2 0 yz 11 2 1 2 2 0 3 K 6 Since the first 3 columns of ( A | b ) are in upper triangular form, | A | 0 and the equations are consistent for all values of K. From the last equation, 3z K 6 z 6K 5 3 3 when K 1 x y z 2 x 2y z 0 is x 4. The same system of equations as in Problem 3, 2z K y a) inconsistent for all values of K b) consistent with solution x 1, y 0, z 1 when K 0 c) consistent with solution x 1, y 0, z 1 when K 3 d) consistent with solution x 1, y 0, z 1 when < K e) none of the above For K 0, 6K 6 2 3 3 For K 3, z z 6 K 63 1 3 3 y 2 z 2 x yz 2 x 1, y 0, z 1 y 2 z - 2 2(1) 2 0 x 2 y z 2 0 1 1 b 5. The system of equations 2c 3 b a c c 2 1 a a) b) c) d) has a unique solution is inconsistent is consistent with one arbitrary unknown none of the above Ax b , 1 1 2 A 1 0 1 , 0 1 1 3 b 2 , 1 a x b c abc 1 1 2 3 ( A | b ) 1 0 1 2 0 1 1 1 1 1 2 3 0 1 1 1 0 1 1 1 1 1 2 3 0 1 1 1 0 0 0 0 The equations are consistent with one arbitrary unknown. 1 1 , 1 2 6. Given matrix A 2 1 1 1 2 1 1 1 c) A 1 1 d) A 1 1 1 2 1 a) A1 does not exist b) A 2 1 f) none of the above 1 1 1 e) A 1 1 A , 1 2 2 1 cof A , 1 1 A1 | A| 11 12 2 1 1 2 1 Adj A (cof A)T 1 1 2 1 1 Adj A | A| 1 1 7. After two iterations using the Jacobi Method to solve 2x y 1 with initial guess x 2 y 1 x0 0, y0 0 results in a) x2 0.75, y2 0.75 b) x2 0.75, y2 0.5 c) x2 0.75, y2 0.75 d) x2 0.5, y2 0.75 e) none of the above 2x y x 2y 1 x 1 y , 2 x0 0 1 y 1 x , 2 y0 0 1 y0 1 0 1 , x1 2 2 2 x2 1 x0 1 0 1 y1 2 2 2 1 y1 1 (0.5) 0.75, 2 2 y2 1 x1 1 (0.5) 0.75 2 2 8. Given the same system of equations in Problem 7 2x y 1 the solution for x is x 1. x 2 y 1 Using the Jacobi Method with initial guess x0 0, y0 0 results in a true error in x, ET x after the second iteration of a) ‐ 0.5 b) 0 c) 0.75 d) 0.25 e) 1 f) none of the above ET x 1 x2 1 0.75 0.25 ...
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