ii) The vectors vi, V2, vs, u and w in R3 are defined by 3 a. 25 2 with the set
ID: 3115289 • Letter: I
Question
ii) The vectors vi, V2, vs, u and w in R3 are defined by 3 a. 25 2 with the set D given by D - (v, v2, vs) and the matrix A defined in the Maple sessiorn below. Yo u may use the following Maple session to assist in answering the questions below. > with(LinearAlgebra): > A:= >; A:=13-5-21 -2725 M:=13-5-21 b -2 725 c > GaussianElimination (M); 1 2 a. 0 -11 -33 b 3a 0 0 0 b+c-a a) Is D a linearly independent set? Give reasons. b) State the condition(s) for the vector w to belong to span(D). c) Determine all possible real scalars 1, 2 and 03 such that d) Find a basis for the kernel of A. e) State the rank of the matrix A. f) Find a basis for the image of A.Explanation / Answer
The RREF of the matrix [A|w] is
1
2
4
a
0
-11
-33
b-3a
0
0
0
b+c-a
(a) Since (4,-33,0)T = 3(2,-11,0)T-2(1,0,0)T, we have v3 = 3v2-2v1 or, 3v2-2v1-v3 = 0. Hence the set D is linearly dependent.
(b) The 3rd row of the RREF of the matrix [A|w] is (0,0,0|b+c-a). Thus if w is in span(D), then b+c-a = 0 or, a = b+c.
(c) Assumig a = 3,b = 1 and c = 2, the RREF of [A|u] is
1
2
4
3
0
-11
-33
-8
0
0
0
0
We may ignore v3 as it is a linear combination of v1 and v2. Now, if u = pv1+qv2, then p+2q = 3 and -11q = -8 or, q = 8/11. Then p +16/11 = 3 so that p = 3-16/11 = 17/11. Thus, u = (17/11)v1+(8/11)v2.
(d) The kernel of A is the set of solutions to the equation AX = 0. If X = (x,y,z)T, then this equationn is equivalent to x+2y +4z = 0, and -11y-33z = 0 or, y+3z = 0 so that y = -3z and x-6z +4z = 0 or, x-2z = 0 so that x = 2z. Then X = (2z,-3z,z)T = z(2,-3,1)T. Thus, Ker(A) = span{(2,-3,1)T }.
(e) The rank of the matrix A, being equal to the number of non-zero rows in its RREF, is 2.
(f) The image of A is same as col(A). Hence a basis for the image of A is { v1,v2}
1
2
4
a
0
-11
-33
b-3a
0
0
0
b+c-a
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