# Thread: Linear Algebra - Fields

1. ## Linear Algebra - Fields

I came across this question in my course.
Consider a system AX=B of n linear equations in n unknowns where A and B have integer entries. Prove or disprove : If the system has an integer solution, then it has a solution in F(p) for all p.

Well, I could see that I have to disprove it. Since the determinant could be 0 (mod p). In which case the system needn't have a solution. But how exactly do i present a formal proof for this?

One more question,
It says in my book for the system AX = B
where
( 8 3 = A
2 6)
and B = (3 -1)t
(I hope you understand the matrix notations.) , there is a solution in F(7), even though det(A)=42 ~0 (mod 7). How is this ?

2. Originally Posted by poorna
I came across this question in my course.
Consider a system AX=B of n linear equations in n unknowns where A and B have integer entries. Prove or disprove : If the system has an integer solution, then it has a solution in F(p) for all p.

Well, I could see that I have to disprove it. Since the determinant could be 0 (mod p). In which case the system needn't have a solution. But how exactly do i present a formal proof for this?
it's actually true for an obvious reason: let $A=[a_{ij}].$ write AX = B as a system of equations: $\sum_{j=1}^n a_{ij}x_j=b_i, \ 1 \leq i \leq n.$ for any integer $a,$ let $\bar{a}$ be $a \mod p.$

then we'll have $\sum_{j=1}^n \bar{a_{ij}} \bar{x_j}=\bar{b_i}, \ 1 \leq i \leq n.$ thus $x=[\bar{x_1} \ \bar{x_2} \ \cdots \ \bar{x_n}]^T$ is a solution modulo p.

One more question,
It says in my book for the system AX = B
where
( 8 3 = A
2 6)
and B = (3 -1)t
(I hope you understand the matrix notations.) , there is a solution in F(7), even though det(A)=42 ~0 (mod 7). How is this ?
$\det A = 0$ does not imply that the equation $AX=B$ has no solution. so there's nothing strange about your example.