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Let $A$ be a non zero $n×n$ invertible matrix with integers entries. Let $b$ a non zero $n×1$ vector with integers entries. Let $x$ be a $n×1$ vectors verifying: $$Ax=b$$

My question is: Show that all the entries of $x$ are rational numbers.

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    $\begingroup$ That isn't necessarily true: let $A$ be the zero matrix and $b$ be the zero vector; then any $x$ satisfies the equation. $\endgroup$ – Solomonoff's Secret Feb 4 '15 at 18:32
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    $\begingroup$ On the other hand, if $A$ is invertible, by the definition of adjugate matrix and determinant, $x$ has to be rational. $\endgroup$ – Troy Woo Feb 4 '15 at 18:33
  • $\begingroup$ @Solomonoff'sSecret: Ok corrected. $\endgroup$ – DER Feb 4 '15 at 18:34
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    $\begingroup$ Alternatively, this follows directly from Cramer's rule. $\endgroup$ – Henning Makholm Feb 4 '15 at 18:34
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    $\begingroup$ What you can prove is that there exists a $x\in\Bbb Q^n$ satisfying the equation. When $A$ is not of maximal rank, $x$ is not unique. $\endgroup$ – AdLibitum Feb 4 '15 at 18:35
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Well now that it's invertible, it's trivial.

$x = A^{-1}b$

$A^{-1}$ cotains just rationals, because of the inverse formula. $A^{-1}b$ contains just sums of products of rationals, thus rational.

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