6252018 Chapter 4 Review Quz 10. Either use an appropriate theorem to show that
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6252018 Chapter 4 Review Quz 10. Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example contrary. s- 3t 2+3s 2s+t 3s :st real The set W is a subset of R If W were a vector space, what else would be true about it? OA. The set W would be a subspace of R OB. The set W would be a subspace of R O C. The set W would be the null space of R4. O D. The set W would be the null space of R2 -31 o Determine whether the zero vector is in W, Find values for t and s such that Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice. The zero vector is in W. The vector equation is satisfied when t and O B. The zero vector is not in W. There is no t and s such that the vector equation is satisfied Which of the following is a true statement? OA. Since the zero vector is not in W. Wis not the null space of R2,Thus Wis not a vector space. O B. Since the zero vector is in W, W is a subspace of R2. Thus W is a vector space. O C. Since the zero vector is in W, W is the null space of R4. Thus W is a vector space. OD. Since the zero vector is not in W, W is not a subspace of R4. Thus W is not a vector space. 11. Find an explicit description of Nul A by listing vectors that span the null space. 1 -50 30 A 0 0 1 -4 0 0 00 05 A spanning set for Nul A is (Use a comma to separate answers as needed.)Explanation / Answer
10. If W were a vector space, then W would be a subspace of R4, as the vectors in W are 4-vectors. Option A is the correct answer.
We have W = { (s-3t,2+3s,2s+t, 3s)T: s, t are real}. Since the equations 2+3s = 0 and 3s = 0 are inconsistent, hence the zero vector (0,0,0,0)T does not belong to W. Therefore, W is not a vector space. Option B is the correct answer.
Since the zero vector (0,0,0,0)T is not in W,W is not a subspace of R4.Thus, W is not a vector space. Option D is the correct answer.
11. The null space of A is the set of solutions to the equation AX = 0. If X = (x, y, z, w, u)T, then this equation is equivalent to x-5y+3w = 0 or, x = 5y-3w, z-4w = 0 or, z = 4w and 5u = 0 or, u = 0 so that X = (5y-3w, y, 4w, w, 0)T = y(5,1,0,0,0)T+w(-3,0,4,1,0)T. Thus, a spanning set for Null(A) is {(5,1,0,0,0)T,(-3, 0, 4,1,0)T }.
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