Suppose I have a $n\times n$ matrix $A$ with integer entries. Suppose $\det(A)=1$.

I was thinking how determinant will change if I reduce the entries of $A$ modulo $k$ where $k$ is a positive integer.

What will be the determinant of this reduced matrix?

I think $\det$ of the reduced matrix will again be $1$.

This is what I have tried so far.

I took a arbitrary $2\times 2$ matrix $$A=\begin{pmatrix}a &b\\c&d\end{pmatrix}$$then we have $ad-bc=1$. Consider the reduced matrix(after going modulo each entry by say $k$)

$$\bar{A}=\begin{pmatrix}\bar{a} &\bar{b}\\\bar{c}& \bar{d}\end{pmatrix}$$ where $\bar{*}:=* \mod k$

Then we are interested in $\bar{a}\bar{d}-\bar{b}\bar{c}$

Now $$ab \mod n = (a \mod n) · (b \mod n)$$ and $$(a + b) \mod n = (a \mod n + b \mod n) \mod n$$

Using these two results we conclude that $\det(\bar{A})=1$

Since $\det$ is a polynomial expression in entries of matrix. We can generalize the process for any $n\times n$ matrix.

Hence in conclusion I propose this lemma.

Let $A=(a_{ij})$ be a $n\times n$ integer matrix with $\det(A)=1$ then consider the matrix $\bar{A}$ whose $ij$ th entry is $a_{ij}\mod k$ where $k$ is a positive integer then we can conclude that $\det(\bar{A})=1$

Am I correct?

  • 3
    $\begingroup$ Yes, it's perfectly correct. $\endgroup$
    – Bernard
    Oct 14, 2018 at 17:06
  • $\begingroup$ Nope. counterexample: $A = [\begin{smallmatrix}4 & 3\\5 & 4\end{smallmatrix}]$ and $k = 5$. $\endgroup$ Oct 14, 2018 at 18:59
  • $\begingroup$ @achillehui It works as $\bar{A }= [\begin{smallmatrix}4 & 3\\0 & 4\end{smallmatrix}]$ and determinant is 16 which is congruent to 1 mod 5 $\endgroup$ Oct 14, 2018 at 19:00

1 Answer 1


You have shown $\overline{ab} = \overline{\overline a \overline b}$ and $\overline{a+b} = \overline{\overline a + \overline b}$. Then you can express $\det(A)$ as in the big formula and obtain $$\overline{\det(\overline A)} = \overline{\sum_{\tau \in S_n}\prod_{i=1}^n\overline{A_{i,\tau(i)}}} = \overline{\sum_{\tau \in S_n}\prod_{i=1}^nA_{i,\tau(i)}} = \overline{\det(A)}$$ with the special case $\overline{\det(A)} = 1 \implies \det(\overline A) = 1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .