(a) The columns of A corresponding to columns of rref(A) that contain a leading 1 (pivot columns of A) give a basis: these are the first and third columns:

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.

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