Bonjour Charline,
[tex]A=\begin{pmatrix}1&3\\0&2\end{pmatrix}\ ;\ B=\begin{pmatrix}2&3\\-1&0\end{pmatrix}\ ;\ C=\begin{pmatrix}-2&1\\1&2\end{pmatrix}[/tex]
[tex]1)\ BC=\begin{pmatrix}2\times(-2)+3\times1&2\times1+3\times2\\(-1)\times(-2)+0\times1&-1\times1+0\times2\end{pmatrix}\\\\\\\boxed{BC=\begin{pmatrix}-1&8\\2&-1\end{pmatrix}}[/tex]
Un calcul analogue montrerait que
[tex]A(BC)=\begin{pmatrix}1&3\\0&2\end{pmatrix}\begin{pmatrix}-1&8\\2&-1\end{pmatrix}\\\\\\\boxed{A(BC)=\begin{pmatrix}5&5\\4&-2\end{pmatrix}}[/tex]
[tex]AB=\begin{pmatrix}1&3\\0&2\end{pmatrix}\begin{pmatrix}2&3\\-1&0\end{pmatrix}\\\\\\\boxed{AB=\begin{pmatrix}-1&3\\-2&0\end{pmatrix}}\\\\\\(AB)C=\begin{pmatrix}-1&3\\-2&0\end{pmatrix}\begin{pmatrix}-2&1\\1&2\end{pmatrix}\\\\\\\boxed{(AB)C=\begin{pmatrix}5&5\\4&-2\end{pmatrix}}[/tex]
Par conséquent,
[tex]\boxed{A(BC)=(AB)C=\begin{pmatrix}5&5\\4&-2\end{pmatrix}}[/tex]
