# $A$ is matrix with integer entries and determinant 1. Effect on determinant if we reduce the entries modulo $k$

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?

• Yes, it's perfectly correct. Oct 14, 2018 at 17:06
• Nope. counterexample: $A = [\begin{smallmatrix}4 & 3\\5 & 4\end{smallmatrix}]$ and $k = 5$. Oct 14, 2018 at 18:59
• @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 Oct 14, 2018 at 19:00

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 $$\overline{\det(A)} = 1 \implies \det(\overline A) = 1$$.