0

Suppose I have a matrix say $A$ with integer elements. I want to have the elements of $A^{-1}$ to be integers . Here $A^{-1}$ stands for inverse matrix of $A$ . How can I prove that this is achievable iff the determinant value of $A$ is unimodular?

  • I have only noticed that if determinant value is +1 or -1 then the situation is true. –  Dec 29 '18 at 17:17
  • 2
    Cramer's rule${}$? – Angina Seng Dec 29 '18 at 17:18
  • Can you please elaborate a little bit @Lord Shark The Unknown? –  Dec 29 '18 at 17:19
  • @Jose I want to prove it for an $n×n$ matrix, the link you have given is only showing the proof for a $2×2$, I guess. –  Dec 29 '18 at 17:28
  • @AkashRoy That's the question, but did you actually read the first answer? – José Carlos Santos Dec 29 '18 at 17:41
  • @Jose Carlos , I have got the answer. But how can I prove it using Cramer's rule as mentioned above by Lord Shark? –  Dec 30 '18 at 10:20
  • I only see how to apply Cramer's rule to prove that if $M$ has integer coefficients and its determinant is $\pm1$, then $M^{-1}$ has integer coefficients, but not in the opposite direction. – José Carlos Santos Dec 30 '18 at 10:34
  • An illustration would have helped me more , but I will try to manage if you can tell me how to start about it. Thanks in advance @Jose –  Dec 30 '18 at 11:21
  • You have $M.M^{-1}=\operatorname{Id}$. Therefore, if $v$ is the first column of $M^{-1}$, then $M.v=(1,0,0,\ldots,0)$. So, $v$ can be computed by Cramer's rule. Since $\det M=\pm1$, $v$ will have integer coefficients. A similar argumen applies to the other columns. – José Carlos Santos Dec 30 '18 at 11:27
  • Thank you @Jose got your approach easily! –  Dec 30 '18 at 11:32

0 Answers0