sample_tt2_1_ans

sample_tt2_1_ans - Math 136 Sample Term Test 2 # 1 Answers...

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Math 136 Sample Term Test 2 # 1 Answers NOTE : - Only answers are provided here (and some proofs). On the test you must provide full and complete solutions to receive full marks. 1. Short Answer Problems a) What is the definition of the row space and column space of a matrix A . Solution: The row space is the spanning set of the rows of A and the column space is the spanning set of the columns of A . b) What is the definition of a subspace. Solution: A subspace is a subset of a vector space V that is also a vector space under the same operations as V . c) What is the definition of a basis. Solution: A linearly independent spanning set. d) Let B = 1 2 1 , - 1 0 2 , 1 1 1 . If [ ~v ] B = 1 - 1 1 what is ~v ? Solution: ~v = 1 1 2 1 + ( - 1) - 1 0 2 + (1) 1 1 1 = 3 3 0 . e) Let V be a vector space and ~v V . Prove that ( - 1) ~v is the additive inverse of ~v . Solution: We have ~v + ( - 1) ~v = (1 + ( - 1)) ~v = 0 ~v = ~ 0 , by V8. Thus ( - 1) ~v is the additive inverse. 2. Let β = { 1 + x 2 , 1 + x + x 2 , 1 + 2 x + 2 x 2 } . a) Show that β is a basis for P 2 . Solution: Consider c 1 (1 + x 2 ) + c 2 (1 + x + x 2 ) + c 3 (1 + 2 x + 2 x 2 ) = 0. Row-reducing the coefficient matrix gives 1 1 1 0 1 2 1 1 2 1 1 1 0 1 2 0 0 1 . Hence, the only solution is c 1 = c 2 = c 3 = 0. Thus β is linearly independent and since the dimension of P 2 is 3 and we have 3 linearly independent vectors in a 3 dimensional space, so it is a basis for P 2 . 1
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2 b) Find the β coordinates of the standard basis vectors of P 2 . Solution: [1] β = 0 2 - 1 , [ x ] β = - 1 1 0 , [ x 2 ] β = 1 - 2 1 . 3. Determine, with proof, which of the following are subspaces of the given vector space. a) S = { ( x 1 ,x 2 ,x 3 ,x 4 ) R 4 ± ± ± ± 2 x 1 - 5 x 4 = 0 and 3 x 2 - 2 x 4 = 0 } of R 4 .
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sample_tt2_1_ans - Math 136 Sample Term Test 2 # 1 Answers...

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