Problem: (1)Consider the following system of equations:

x1 + x2 + x3 = 2

2x1 + x2 + 3x3 = 4

6x1 + 2x2 + 3x3 = 5

3x1 + 2x2 + 4x3 = 6

a) Viewing the system as a system equations over R (the field of real numbers), write down the augmented matrix

b) Find the reduced row echelon form (RREF) of the augmented matrix indicating your steps.

c) Recover an equivalent system from the RREF and give the solution set to the system.

(2) Repeat problem 1 except this time view the system as system of equations over the field

Z7 = {0, 1, 2, 3, 4, 5, 6} mod 7.

Attmpt to the solution.This is my solution to question 1:

(a)

1 1 1 | 2

2 1 3 | 4

6 3 2 | 5

3 2 4 | 6

(b)

R2 -> R2 – 2 R1

1 1 1 | 2

0 -1 1 | 0

6 3 2 | 5

3 2 4 | 6

R3 -> R3 – 6 R1

1 1 1 | 2

0 -1 1 | 0

0 -3 -4 | -7

3 2 4 | 6

R4 -> R4 – 3 R1

1 1 1 | 2

0 -1 1 | 0

0 -3 -4 | -7

0 -1 1 | 0

R2 -> -1 * R2

1 1 1 | 2

0 1 -1 | 0

0 -3 -4 | -7

0 -1 1 | 0

R3 -> R3 +3 R2

1 1 1 | 2

0 1 -1 | 0

0 0 -7 | -7

0 -1 1 | 0

R4 -> R4 + R2

1 1 1 | 2

0 1 -1 | 0

0 0 -7 | -7

0 0 0 | 0

R3 -> R3/-7

1 1 1 | 2

0 1 -1 | 0

0 0 1 | 1

0 0 0 | 0

R2 -> R2 + R3

1 1 1 | 2

0 1 0 | 1

0 0 1 | 1

0 0 0 | 0

R1 -> R1 - R3

1 1 0 | 1

0 1 0 | 1

0 0 1 | 1

0 0 0 | 0

R1 -> R1 – R2

1 0 0 | 0

0 1 0 | 1

0 0 1 | 1

0 0 0 | 0

This augmented matrix is in RREF.

(c) Equivalent system:

x1 + 0x2 + 0x3 = 0

0x1 + x2 + 0x3 = 1

0x1 + 0x2 + x3 = 1

0x1 + 0x2 + 0x3 = 0

Solution set:

x1 = 0

x2 = 1

x3 = 1

But I do not know how to answer question 2.