Determine whether the set of all third-degree polynomial functions as given belo
ID: 3038017 • Letter: D
Question
Determine whether the set of all third-degree polynomial functions as given below, with the standard operations, is a vector space. If it is not, then determine the set of axioms that it fails. ax^3 + bx^2 + cx + d, a notequalto 0 Verity W = {(x, y, 3x-8y): x and y are real numbers} is a subspace of V = R^2. Determine whether the set S = {(-5, 6), (3, 7)} is linearly independent or linearly dependent. Determine whether the set S = {(3, 1), (-1, 3)} spans R^2. Explain why S = {(6, -5), (12, -10)} is not a basis for R^2. Find the rank of the matrix [-4 3 0 -24 18 0 24 -18 0 11 -5 1]. Find a basis for the subspace of R^3 spanned by S = {(16, 4, 30), (8, 2, 15), (4, 1, 5)}. Find the length of the vector v = (2, 4, 2). Find the distance d between u = (3, 3, 6) and v = (-3, 3, 0). Find the angle theta between the vectors u = (0, 5, 0, 5) and v = (3, 2, 7, 3). Find the vector v with length 3 and the same direction as the vector u = (-5, 5, 1).Explanation / Answer
27. Let X = ax3+bx2 +cx+d and Y = px3+qx2 +rx +s ( a 0 and p0) be 2 arbitrary elements of S, the set of all 3rd degree polynomial functions and let k be an arbitrary scalar.Then X+Y=(ax3+bx2+cx+d)+(px3+qx2+rx +s) = (a+p)x3+(b+q)x2+ (c+r)x+(d+s). Now, if p = -a, then a+p = 0. In such a case, X+Y S. Thus, S is not closed under vector addition, and hence S is not a vector space).
28. Here, V = {(x,y,3x-8y): x, y R}. Let X = (x, y,3x-8y) and Y = (z,w,3z-8w) be 2 arbitrary elements of V and let k be an arbitrary scalar. Then X+Y=( x, y,3x-8y)+ (z,w,3z-8w)=(x+z, y+w, 3x+3z-8y-8w). Also, 3x+3z=8y-8w = 3(x+z)-8(y+w) so that X+Y V. Therefore, V is closed under vector addition. Further, kX = k(x, y, 3x-8y) = [kx, ky, k(3x-8y)] = (kx,ky, 3kx -8ky). Hence kX V . This means that V is closed under scalar multiplication. Also, apparently, the zero vector (0,0,0) V. Hence V is a vector space, and therefore, a subspace of R3.
29. We have S = {(-5,6),(3,7)}. Let us assume that S is linearly dependent. Then, there are scalars a and b, not both zero, such that a(-5,6)+b(3,7) = 0 or, (-5a,6a)+(3b,7b) = 0 or, (-5a+3b, 6a+7b) = 0. Then, -5a +3b = 0 or, 5a = 3b or, a = 3b/5. Also, 6a+7b = 0 so that 6a= -7b and a = -7b/6. Then a = 3b/5 = -7b/6. This is, however possible only when b = 0 (and a=0). This is a contradiction. Hence S is linearly independent.
30. Here, S = {(3,1), (-1,3). Let A =
3
-1
1
3
We will reduce A to its RREF as under:
Multiply the 1st row by 1/3 ; Add -1 times the 1st row to the 2nd row
Multiply the 2nd row by 3/10; Add 1/3 times the 2nd row to the 1st row
Then the RREF of A is I2 . This means that The vectors in S span R3.
31. Here S = {(6,-5),(12,-10)}. Since (12,-10)= 2(6,-5), hence the vectors in S are linearly dependent and S cannot be a basis for R2.
32. Let A =
-4
-24
24
11
3
18
-18
-5
0
0
0
1
We will reduce A to its RREf as under:
Multiply the 1st row by -1/4; Add -3 times the 1st row to the 2nd row
Multiply the 2nd row by 4/13; Add -1 times the 2nd row to the 3rd row
Add 11/4 times the 2nd row to the 1st row
Then, the RREF of A is
1
6
-6
0
0
0
0
1
0
0
0
0
The rank of A, being the number of non-zero rows in its RREF is 2.
Please post the rtemaining questions again, mmaximum 4 at a time.
3
-1
1
3
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