Thread: Augmented matrix problem: when field of system of equations is given

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.

Why 4 equations; you only need 3.

Suggest you use a,b,c instead of x1,x2,x3; easier to work with....

Do you know what "modulo 7" means? Members of are "equivalence classes" of numbers having the same remainder when divided by 7. That is, a= b (mod 7) if a= b+ 7m for some integer m. Typically, we represent each equivalence class by that remainder which must be one of 0, 1, 2, 3, 4, 5, 6, or 7. For example 5+ 4= 9= 7+ 2 so "5+ 4= 2 (mod 7)". Also 3(6)= 18= 2(7)+ 4 SO "3(6)= 4 (mod 7)". And, 3+ 4= 7= 0 (mod 7) so -3= 4 (mod 7). When you "subtract twice the first row from the second row", every number, except in the second column, involved is between 0 and 6 so there is no difference. In the second column, 1- 2= -1= 6 (mod 7). But when you "subtract 6 times the first row from the third row", 3- 6= -3= 4 (mod 7), 2- 6= -4= 3 (mod 7) and 5- 6(2)= -7= 0 (mod 7) so your third rwo becomes (0 4 3 0) rather than (0 -3 -4 -7).

Simlarly, when you subtract three times the first row from the fourth row, 3- 3= 0, 2- 3= -1= 6, 4- 1= 1 and 6- 6= 0 (mod 7)so after "reducing" the first column you will have