### Autor Tema: Inverse of a integer matrix

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09 Abril, 2022, 08:34 pm
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#### Kandor

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##### Inverse of a integer matrix
Let
$$M:=\left\{\left(\begin{array}{ll} a & b \\ 0 & c \end{array}\right) \mid a, b, c \in \mathbb{Z}\right\} .$$
$$M$$ is a non-commutative monoid under matrix multiplication. The element $$\left(\begin{array}{ll}a & b \\ 0 & c\end{array}\right)$$ has an inverse if and only if $$a, c \in\{\pm 1\}$$. In this case, one has
$$\left(\begin{array}{ll} a & b \\ 0 & c \end{array}\right)^{-1}=\left(\begin{array}{cc} a & -a b c \\ 0 & c \end{array}\right)$$
(Verify!)

09 Abril, 2022, 11:43 pm
Respuesta #1

#### Luis Fuentes

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##### Re: Inverse of a integer matrix
Hi

Let
$$M:=\left\{\left(\begin{array}{ll} a & b \\ 0 & c \end{array}\right) \mid a, b, c \in \mathbb{Z}\right\} .$$
$$M$$ is a non-commutative monoid under matrix multiplication. The element $$\left(\begin{array}{ll}a & b \\ 0 & c\end{array}\right)$$ has an inverse if and only if $$a, c \in\{\pm 1\}$$. In this case, one has
$$\left(\begin{array}{ll} a & b \\ 0 & c \end{array}\right)^{-1}=\left(\begin{array}{cc} a & -a b c \\ 0 & c \end{array}\right)$$
(Verify!)

Hace you tried something?

Since the matrix product is associative with identity element the indentity matrix, it's sufficiente to check that the multiplication restricts to $M$. But:

$\begin{pmatrix}a&b\\0&c\\\end{pmatrix}\begin{pmatrix}a'&b'\\0&c'\\\end{pmatrix}=\begin{pmatrix}aa'&ab'+bc'\\0&cc'\\\end{pmatrix}$

Moreover, if $\begin{pmatrix}a&b\\0&c\\\end{pmatrix}$ has inverse $\begin{pmatrix}a'&b'\\0&c'\\\end{pmatrix}$ then:

$\begin{pmatrix}aa'&ab'+bc'\\0&cc'\\\end{pmatrix}=\begin{pmatrix}1'&0\\0&1\\\end{pmatrix}$

From this $aa'=1$ and $cc'=1$. But $a,a',c,c'$ are integers so $a'=a=\pm 1$, $c'=c=\pm 1$ and:

$ab'+bc'=0\quad \Rightarrow \quad ab'+bc=0 \quad \Rightarrow \quad ab'=-bc \quad \Rightarrow \quad aab'=-abc \quad \Rightarrow \quad b'=-abc$