# Thread: solve an xor'd linear equation

1. ## solve an xor'd linear equation

hello,
i have 6 variables
a=5
b=10
c=7
d=9
x=1
y=2
when i solve this equation using gauss-jordan it gives me wrong answer
a⊕0⊕c⊕0⊕0⊕0=2
0⊕b⊕0⊕d⊕0⊕0=3

a⊕0⊕0⊕0⊕x⊕0=4
0⊕b⊕0⊕0⊕0⊕y=8

0⊕0⊕c⊕0⊕x⊕0=6
0⊕0⊕0⊕d⊕0⊕y=11

⊕ is XOR

2. ## Re: solve an xor'd linear equation

The initial augmented matrix is
Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
1 0 0 0 1 0 = 4
0 1 0 0 0 1 = 8
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
Replace row 3 by the sum of row 1 and row 3 (in the future, 3 := 1 + 3).

Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
0 0 1 0 1 0 = 6
0 1 0 0 0 1 = 8
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
4 := 2 + 4

Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
5 := 3 + 5

Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
0 0 0 0 0 0 = 0
0 0 0 1 0 1 = 11
6 := 4 + 6

Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
0 0 0 0 0 0 = 0
0 0 0 0 0 0 = 0
The last two equations are superfluous

Code:
1 0 1 0 0 0 = 2
0 1 0 1 0 0 = 3
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
2 := 4 + 2

Code:
1 0 1 0 0 0 = 2
0 1 0 0 0 1 = 8
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
1 := 3 + 1

Code:
1 0 0 0 1 0 = 4
0 1 0 0 0 1 = 8
0 0 1 0 1 0 = 6
0 0 0 1 0 1 = 11
This is a reduced row echelon form. Given the values of x and y we can find a, b, c and d. In particular, for x = 1 and y = 2,

d + 2 = 11; d = 9
c + 1 = 6; c = 7
b + 2 = 8; b = 10
a + 1 = 4; a = 5