1
2
3

and

2
5
7

The dimension is 2.

[Note that taking the first and third columns from rref(A) gives the wrong answer: the linear combinations of those all have a 0 in the third entry, which is not true of all vectors in the column space if A.]

(b) Using the rref to solve for leading variable in terms of free variables (say we use v, w, x, y, z) gives as the general solution:

     -2w - 14y -23z
                 w
{             5y + 9z        :        w, y, z in R }
                 y
                 z

(v and x are leading variables; w, y, z are free)
and since the general solution vector can be re-written

     -2              -14            -23
       1                  0                0
w     0    +    y    5    +    z    9
       0                  1                0
       0                   0                1

the three vectors

-2
   1
   0
   0
   0 ,

-14
     0
     5
     1
     0 ,

and

-23
     0
     9
     0
     1

are a basis for the kernel. The dimension of the kernel is 3.

2.
The dimension of the kernel is the dimension of the domain minus the dimension of the image: in this case, 7 - 4 = 3.

3.
We need that the dot product be zero, or 3(-1) + 2(-2) + 1(k) = 0, i.e., -3 -4 + k = 0. Thus, k = 7 is the only value that produces perpendicular vectors.