For the given matrix A and vector b select the true statemen
Solution
Let B = [ A|b] =
1
5
0
1
3
5
1
5
1
2
0
4
2
1
1
The RREf of B is
1
0
0
3/68
49/68
0
1
0
13/68
31/68
0
0
1
2/17
-7/17
Then,
1.dim( Col(A)) = 3;
2. Null(A) being the set of solutions to the equation AX = 0, is a subspace of R4;
3.b is a 3-vector and Null(A) contains 4-vectors so that b cannot be in Null(A);
4.Col(A) is all of R3 as its columns are linear combinations of (1,0,0)T, (0,1,0)T and (0,0,1)T;
5.If X = (x,y,z,w)T, then Null(A) is the set of solutions to the equations x +3w/68 = 0, y +13w/68 = 0 and z +2w/17 = 0 so that X = (-3w/68,-13w/68, -2w/17,w)T = w/68( -3,-13,8,68)T. Thus Null(A) = span{( -3,-13,8,68)T } so that dimension of Null(A) is 1;
6. b = (49/68)(1,5,0 )T + (31/68)(5,1,4 )T-(7/17)(0,5,2 )T so that b is in Col(A).
The statements numbers 4,5, and 6 are true.
| 1 | 5 | 0 | 1 | 3 |
| 5 | 1 | 5 | 1 | 2 |
| 0 | 4 | 2 | 1 | 1 |
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