Pour \( n=1 ), on a ncessairement \( {\rm det }(a)=a ). 


Pour \(n=2), on a ncessairement 
<center>det \(\begin{pmatrix}\special{color=green}a\special{color=black}&b\\
\special{color=green}c\special{color=black}&d\end{pmatrix})= det \(\begin{pmatrix}\special{color=red}a\special{color=black}+
\special{color=orange}0\special{color=black}&b\\\special{color=red}0\special{color=black}
+\special{color=orange}c\special{color=black}&d\end{pmatrix})= 
det \(\begin{pmatrix}\special{color=red}a\special{color=black}
&b\\\ \special{color=red}0\special{color=black}&d\end{pmatrix})+ 
det \(\begin{pmatrix}\special{color=orange}0\special{color=black}&b\\ \special{color=orange}c\special{color=black}&d\end{pmatrix})</center>
<center>
= \(a) det \( \begin{pmatrix}1&\special{color=green}b\special{color=black}\\0&\special{color=green}d\special{color=black}\end{pmatrix}) + c det\(\begin{pmatrix}0&\special{color=green}b\special{color=black}\\1&\special{color=green}d\special{color=black}\end{pmatrix}) 
</center>
<center>
= \(a) ( det \( \begin{pmatrix}1&\special{color=red}b\special{color=black}\\0&\special{color=red}0\special{color=black}\end{pmatrix}) + det \( \begin{pmatrix}1&\special{color=orange}0\special{color=black}\\0&\special{color=orange}d\special{color=black}
\end{pmatrix})) + c (det \(\begin{pmatrix}0&\special{color=red}b\special{color=black}\\1&\special{color=red}0\special{color=black}\end{pmatrix}) +det\(\begin{pmatrix}0&\special{color=orange}0\special{color=black}\\ 1&\special{color=orange}d\special{color=black}\end{pmatrix}))
</center>
<center>
= \(a ( 0 + d)) det \( \begin{pmatrix}1&0\\0&1\end{pmatrix}) + \(c) (\(b) det \(\begin{pmatrix}0&1\\1&0\end{pmatrix}) + 0)
= \(ad -bc)</center>