System of linear equations with whole numbers

Question

I need some advice regarding this question.

"Show that if in the system $Ax=b$, the determinant of $A$ is equal to $-1$, and all the members of $A$ are whole numbers, and all the members of $b$ are whole numbers, then the system has single solution, and all it's members are whole numbers."

Thank you in advance.

Answer 1

I guess that should be "integers" instead of "whole numbers".

det $A=-1\neq 0$. So A is invertible. Hence, $x=A^{-1}b$.

$A^{-1}=\frac{Adj A}{det A}$. Since $A$ has integers, Adj $A$ has integers and if we divide them by $-1$ still they are integers. Hence $A^{-1}b$ also has integers.

Answer 2

The solution of the system $Ax=b$ is unique given $\det A \neq 0$ for any real entry matrix $A$ and column vector $b.$ Furthermore, an explicit expression for the solution may be obtained in this case by Cramer's rule, which says each entry of the solution vector $x$ is a fraction with denominator $\det A$ and numerator obtained by forming the determinant of $A$ after one of its columns is replaced by the column vector $b$. So under your assumptions, these denominators are each $-1$ and the numerators are integers.