Question

1 23 -3 6 7 1 1 2-1 2 4 2 24-2 4 8 (a) Determine the vectors that are in the solution space of A (b) What is the rank of A? (c) What is the dimension of the solution space of A? (d) Determine the vectors that are in the column space of A 13 10

appreciate a clear explanation .tks (thumbup)

Answer 1

(a). The RREF of A is

1	0	-1	0	-1	-3
0	1	1	0	-2	-1
0	0	0	1	-3	-2
0	0	0	0	0	0

Now, if X = ( x,y,z,w,u,v)^T, then the solution set of A, i.e. the solutions to the equation AX = 0 satisfy x-z-u-3v = 0 or, x = z+u+3v, y+z-2u-v = 0 or, y = - z+2u+v and w-3u-2v = 0 or, w = 3u+2v.

Then X = (z+u+3v, - z+2u+v, z, 3u+2v,u,v)^T = z(1,-1,1,0,0,0)^T +u(1,2,0,3,1,0)^T +v(3,2,0,2,0,1)^T. This implies that every vector in the solution space of A is a linear combination of 3 linearly independent vectors (1,-1,1,0,0,0)^T , (1,2,0,3,1,0)^T , (3,2,0,2,0,1)^T. Hence the set {(1,-1,1,0,0,0)^T , (1,2,0,3,1,0)^T , (3,2,0,2,0,1)^T } is a basis for the solution space of A.

Now, let M =

1	1	3	4	2	-2	8
-1	2	2	-3	0	0	-2
1	0	0	2	2	3	1
0	3	2	-1	1	1	0
0	1	0	-1	-3	-7	-2
0	0	1	1	-2	6	3

The RREF of M is

1	0	0	0	0	0	-5
0	1	0	0	0	0	1
0	0	1	0	0	0	0
0	0	0	1	0	0	3
0	0	0	0	1	0	0
0	0	0	0	0	1	0

This implies that none of the vectors u₁,u₂,u₃,u₄ can be expressed as a linear combination of the vectors (1,-1,1,0,0,0)^T , (1,2,0,3,1,0)^T , (3,2,0,2,0,1)^T. Hence none of the vectors u₁,u₂,u₃,u₄ is in the solution space of A.

(b). The RREF of A has 3 non-zero rows. Therefore, rank(A) = 3.

( c). The dimension of the solution space of A is also 3 ( equal to the no. of vectors in its basis).

(d). It may be observed from the RREF of A that only its 1^st, 2^nd and 4^th columns are linearly independent.

Now, let N =

-1	2	-3	2	13	-4	-3
1	-1	2	3	-8	0	2
-1	1	-1	-1	5	2	-1
-2	2	-2	-5	10	4	-2

The RREF of N is

0	0	0	0	0	-6	0
0	1	0	0	2	-2	0
0	0	1	0	-3	2	1
0	0	0	1	0	0	0

This implies that the vectors v₂,v₃,v₄ can be expressed as linear combination of the columns of A. Therefore, the vectors v₂,v₃,v₄ are in the column space of A.