For typographical reasons, brackets around matrices are omitted here.

1.
The matrix of the system is

1 5 0 4
3 0 1 7

A dotted vertical line separating the fourth (rightmost) column is optional. (As mentioned in class, this problem was worth 2, not 3 points.)

2.
One can reach reduced row echelon form in two steps. Divide the first row by 2, giving

1 2 3 4
0 0 1 1

Now subtract 3 times the second row from the first, which gives the answer:

1 2 0 1
0 0 1 1

The size of the matrix is 2 by 4.

3.
Here, x_1 means "x with the subscript 1", etc. The solutions are found by solving for the leading variables (x_1 and x_3) in terms of the TWO free variables (x_2 and x_4). This gives

x_1 = 8 - 3x_2 - 2x_4
x_3 = 3 - x_4

The solution set is the set of all vectors of the form

8 - 3x_2 - 2x_4
x_2
3 - x_4
x_4

with x_2 and x_4 in R = the real numbers (the vertical square brackets have been omitted from the description of this column vector) or, in set notation,

         8 - 3x_2 - 2x_4
{        x_2                                   : x_2, x_4 in R }
         3 - x_4
         x_4

(again, the vertical square brackets around the column vector have been omitted). There was a small penalty if the solution set was not presented AS A SET.