Linear Algebra : Fields, Matrices

• Jul 24th 2009, 09:18 AM
Linear Algebra : Fields, Matrices
I'm in the subject 'Matrices', but there still is a use of the Fields topic.

I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

Now, I thought of the following system:

3x+7y=14
6x+14y=28

x,y(belong to)Z7

Z7 is the following field:
(0,1,2,3,4,5,6)

By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

(0,0), (0,1), (0,2), ... (0,6)
(1,0) . . . (1,6)
.
.
.
.
(6,0), . . . . . (6,6)

and so on...

Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

Thank you :)
• Jul 29th 2009, 08:16 AM
ynj
Quote:

I'm in the subject 'Matrices', but there still is a use of the Fields topic.

I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

Now, I thought of the following system:

3x+7y=14
6x+14y=28

x,y(belong to)Z7

Z7 is the following field:
(0,1,2,3,4,5,6)

By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

(0,0), (0,1), (0,2), ... (0,6)
(1,0) . . . (1,6)
.
.
.
.
(6,0), . . . . . (6,6)

and so on...

Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

Thank you :)

if there are m variables and the rank of these equations(or matrix) is r,then there is 7^(m-r) solutions.....
• Jul 29th 2009, 10:08 AM
HallsofIvy
Quote:

I'm in the subject 'Matrices', but there still is a use of the Fields topic.

I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

Now, I thought of the following system:

3x+7y=14
6x+14y=28

x,y(belong to)Z7

Z7 is the following field:
(0,1,2,3,4,5,6)

By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

(0,0), (0,1), (0,2), ... (0,6)
(1,0) . . . (1,6)
.
.
.
.
(6,0), . . . . . (6,6)

and so on...

Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

I'm not sure what you mean by that. Certainly $Z_7\times Z_7$ contains exactly 49 members but what makes you think they are all solutions to the equations? You mention (1,0) but if x= 1, y= 0, the equations become 3(1)+ 7(0)= 14 and 6(1)+ 14(0)= 28 which are not true, even in $Z_7$.
In fact, in $Z_7$, 7, 14, and 48 are all congruent to 0, so your equations become just 3x= 0 and 6x= 0 (mod 7). While y can be any number, x must be 0. There are only 7 solutions, not 49.
How about the system $x=x$ over the field with 49 elements? (Nod)