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129a-2_test1_sol

# 129a-2_test1_sol - = 1 0 0 0 1 0 0 0 1 has a pivot in the...

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Spring 2010 MATH129A Linear Algebra I, section 2 MW 13:30 – 14:45 TEST 1 Solution 1. (a) False (b) False (c) False (d) False (e) False 2. (a) List those are inconsistent: (b), (e) List those are having inﬁnite solutions: (a), (d) (b) The general solution in parametric vector form is 6 0 7 0 8 + x 2 - 4 1 0 0 0 + x 4 - 3 0 - 5 1 0 . 3. (a) x 1 = 1 ,x 2 = 0 ,x 3 = 2 ,x 4 = 5 ,x 5 = - 1 ,x 6 = 9 ,x 7 = 1 ,x 8 = 2 ,x 9 = - 1 ,x 10 = 33 (b) Take a = 29 4 ,b = - 7 4 ,c = 9 4 . Then A 1 a b c = A 1 29 4 - 7 4 9 4 = A 3 4 3 9 - 2 0 + 1 4 - 5 2 - 1 9 = 3 4 A 3 9 - 2 0 + 1 4 A - 5 2 - 1 9 = 3 4 b + 1 4 b = b 4. (a) NO, the vectors are not linearly independent because rref 2 3 5 - 5 1 - 4 - 3 - 1 - 4 1 0 1 = 1 0 1 0 1 1 0 0 0 0 0 0 does not have a pivot in every column. (b) NO, the vector is not in the span because rref 1 0 4 0 3 0 - 2 6 0

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Unformatted text preview: = 1 0 0 0 1 0 0 0 1 has a pivot in the last column. 1 (c) No value of h will make the vectors linearly independent. 5. (a) T fails (i) because " 2 2 # = T " 1 2 #! = T " 1 1 # + " 1 #! 6 = T " 1 1 #! + T " 1 #! = " 1 1 # + " 1 # = " 1 2 # T fails (ii) because " 1-1 # = T "-1-1 #! 6 = (-1) T " 1 1 #! = (-1) " 1 1 # = "-1-1 # (b) YES, T is onto. (c) "-7 22 # 6. (a) -2 2 (b) NO, because all vectors in Col(A) have zero last component. (c) NO, because A 3-2 1 = 18 , which is not a zero vector. (d) 3 2-1 2...
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129a-2_test1_sol - = 1 0 0 0 1 0 0 0 1 has a pivot in the...

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