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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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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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:

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

Complete details. Ask your doubts.

Best regards.