# UW Common Math 308 Section 4.pdf - JACOB LEE BERNARDI Math...

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Current Score : 29 / 30 Due : Tuesday, February 21 2017 11:00 PM PST 1. 1/1 points | Previous Answers HoltLinAlg1 4.1.003. Determine if the described set is a subspace. Assume a and b are real numbers. The subset of consisting of vectors of the form where If so, give a proof. If not, explain why not. 0+0 doesn't equal 1 R 2 , a b a + b = 1. The set is a subspace. The set is not a subspace. Key: Answers will vary. Score: 0 out of 0 Comment: Solution or Explanation Not a subspace, because and thus 0 is not in this set. UW Common Math 308 Section 4.1 (Homework) JACOB LEE BERNARDI Math 308, section L, Winter 2017 Instructor: Harishchandra Ramadas TA WebAssign The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension 0 + 0 1,
2. 1/1 points | Previous Answers HoltLinAlg1 4.1.006. Determine if the described set is a subspace. Assume a and b are real numbers. The subset of consisting of vectors of the form If so, give a proof. If not, explain why not. 0 is in s u is in s, v is in s, and u+v is in s too r is in s, u is in s, and ru is in s too R 4 . a a + 6 a b 9 b The set is a subspace. The set is not a subspace. Key: Answers will vary. Score: 0 out of 0 Comment: Solution or Explanation Let S be the set of vectors of the form Then S = span and hence S is a subspace. b . a a + b 6 a b 9 b , , 1 1 6 0 0 1 1 9
3. 0/1 points | Previous Answers HoltLinAlg1 4.1.007. Determine if the described set is a subspace. Key: Answers will vary. Score: 0 out of 0 Comment: .
4. 1/1 points | Previous Answers HoltLinAlg1 4.1.009. Determine if the described set is a subspace. Assume a , b , and c are real numbers. The subset of consisting of vectors of the form where If so, give a proof. If not, explain why not. u=(1,0,1) and v=(1,1,0) Key: Answers will vary. Score: 0 out of 0 Comment: R 3 , a b c abc = 0. The set is a subspace. The set is not a subspace. . =