$\begin{aligned} & A=\left(\begin{array}{ccc} 3 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right) \\ & A^2=A \cdot A=\left(\begin{array}{ccc} 3 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right) \cdot\left(\begin{array}{ccc} 3 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right)=\left(\begin{array}{ccc} 5 & 4 & -4 \\ -2 & -1 & 2 \\ 2 & 2 & -1 \end{array}\right) \\ & A^3=A^2 \cdot A=\left(\begin{array}{ccc} 5 & 4 & -4 \\ -2 & -1 & 2 \\ 2 & 2 & -1 \end{array}\right) \cdot\left(\begin{array}{ccc} 3 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right)=\left(\begin{array}{ccc} 7 & 6 & -6 \\ -3 & -2 & 3 \\ 3 & 3 & -2 \end{array}\right) \\ & A^3-3 \cdot A^2+3 \cdot A=\left(\begin{array}{ccc} 7 & 6 & -6 \\ -3 & -2 & 3 \\ 3 & 3 & -2 \end{array}\right)-3\left(\begin{array}{ccc} 5 & 4 & -4 \\ -2 & -1 & 2 \\ 2 & 2 & -1 \end{array}\right)+3\left(\begin{array}{ccc} 3 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I_3 \end{aligned}$ On en déduit : $\begin{gathered} A^3-3 \cdot A^2+3 \cdot A=I_3 \Leftrightarrow A \cdot \underbrace{\left(A^2-3 \cdot A+3 \cdot I_3\right)}_{=A^{-1}}=I_3 \\ \Rightarrow A^{-1}=A^2-3 \cdot A+3 \cdot I_3 \end{gathered}$

Inversibilité de matrices