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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 Spring '11
 klee
 Equations, Emoticon

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