1. ## Integer Valued Matrix

Problem( $\star\star$): Let $A,B\in\text{Mat}_2\left(\mathbb{Z}\right)$ be such that $A,A+B,A+2B,A+3B$ and $A+4B$ are invertible and such that $A^{-1},\left(A+B\right)^{-1},\left(A+2B\right)^{-1},\left(A+3B\right)^{-1},\left(A+4B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right)$. Prove that $A+5B$ is invertible and $\left(A+5B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right)$.

Remark: If you have seen this problem before, or particularly where it comes from, please resist posting.

2. Originally Posted by Drexel28
Problem( $\star\star$): Let $A,B\in\text{Mat}_2\left(\mathbb{Z}\right)$ be such that $A,A+B,A+2B,A+3B$ and $A+5B$ are invertible and such that $A^{-1},\left(A+B\right)^{-1},\left(A+2B\right)^{-1},\left(A+3B\right)^{-1},\left(A+4B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right)$. Prove that $A+5B$ is invertible and $\left(A+5B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right)$.

Remark: If you have seen this problem before, or particularly where it comes from, please resist posting.

Hint:

Spoiler:

First try proving that if $A\in\text{Mat}_n\left(\mathbb{Z}\right)$ then a necessary and sufficient condition for $A$ to be invertible and $A\in\text{Mat}_n\left(\mathbb{Z}\right)$ is that $\det A=\pm 1$

3. Since no one seems interested I'll just give the answer for logical soundness:

Spoiler:

To solve this we first make the observation that if $A\in\text{Mat}_n\left(\mathbb{Z}\right)$ then $A$ is invertible and $A^{-1}\in\text{Mat}_n\left(\mathbb{Z}\right)$ if and only if $\det A=\pm 1$. Indeed, clearly if $\det A=\pm 1$ the result follows. Conversely, the conclusion follows by noticing that since $A^{-1}\in\text{Mat}_n\left(\mathbb{Z}\right)$ that $\det A^{-1}=\frac{1}{\det A}\in\mathbb{Z}$ from where the conclusion follows.

So, with this observation let $p(x)=\det (A+xB)$. Then, since $A,B\in\text{Mat}_2\left(\mathbb{Z}\right)$ we see that $\deg p\leqslant 2$. Note though that by assumption of the problem and the above observation we have that $p(x)\in\{-1,1\},\quad x\in\{0,1,2,3,4\}$. But, by the Pigeon Hole principle we know that $p(x)$ takes the value either $1$ or $-1$ at least three times on $\{0,1,2,3,4\}$. But, since $\deg p\leqslant 2$ this implies that $p$ is constant and $p(x)=1$ or $p(x)=-1$. Thus, by our observation we may conclude that $A+xB$ is invertible and $\left(A+xB\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right)$ for every $x\in\mathbb{Z}$. $\blacksquare$