\begin{enumerate}
\item On a :
\[
\begin{array}{l@{~=~}l}
AI & 
\begin{pmatrix}-1 & 2\\1 & -1\end{pmatrix}
\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\\
& 
\begin{pmatrix}
-1\times1+2\times0 & -1\times0+2\times1\\
1\times1+(-1)\times0 & 1\times0+(-1)\times1
\end{pmatrix}\\
AI & \begin{pmatrix}
-1 & 2\\
1 & -1
\end{pmatrix}\\
AI & A
\end{array}
\]
De plus,
\[
\begin{array}{l@{~=~}l}
IA & 
\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
\begin{pmatrix}-1 & 2\\1 & -1\end{pmatrix}\\
& 
\begin{pmatrix}
1\times(-1)+0\times1 & 1\times2+0\times(-1)\\
0\times1+1\times1 & 0\times2+1\times(-1)
\end{pmatrix}\\
IA & \begin{pmatrix}
-1 & 2\\
1 & -1
\end{pmatrix}\\
IA & A
\end{array}
\]
\uline{On a donc bien $AI=IA=A$.}

\item $\begin{aligned}[t]
AB=I & \iff \begin{pmatrix}-1 & 2\\1 & -1\end{pmatrix}\begin{pmatrix}x & y\\z & t\end{pmatrix}=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\\
& \iff \begin{pmatrix}-x+2z & -y+2t\\x-z & y-t\end{pmatrix}=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\\
& \iff \left\{\begin{array}{l}
-x+2z=1\\
x-z=0
\end{array}\right.\quad\text{et}\quad \left\{\begin{array}{l}
-y+2t=0\\
y-t=1
\end{array}\right.\\
& \iff \left\{\begin{array}{l}
-x+2x=1\\
x=z
\end{array}\right.\quad\text{et}\quad \left\{\begin{array}{l}
y=2t\\
2t-t=1
\end{array}\right.\\
& \iff x=1,\ z=1,\ t=1,\ y=2.
\end{aligned}$

Ainsi, $\boxed{B=\begin{pmatrix}1 & 2\\1 & 1\end{pmatrix}}$.

\item $\begin{aligned}[t]
BA & = \begin{pmatrix}1 & 2\\1 & 1\end{pmatrix}
\begin{pmatrix}-1 & 2\\1 & -1\end{pmatrix}\\
BA & = \begin{pmatrix}
1\times(-1)+2\times1 & 1\times2+2\times(-1)\\
1\times(-1)+1\times1 & 1\times2+1\times(-1)
\end{pmatrix}\\
BA & = \begin{pmatrix}
1 & 0\\0 & 1
\end{pmatrix}\\
\Aboxed{BA & = I}
\end{aligned}$
\end{enumerate}