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Trouver A⁻¹, l'inverse de A. La matrice donnée est $A = eginpmatrix 2 & 1 & 1 \ 1 & 3 & -1 \ 1 & 1 & 2

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5. (a)(i) Trouver A−1A^{-1}, l'inverse de A.
La matrice donnée est A=(21113−1112)A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & 1 & 2 \end{pmatrix}.

  1. Calculer le déterminant de A, det⁡(A)\det(A).

    det⁡(A)=2∣3−112∣−1∣1−112∣+1∣1311∣\det(A) = 2\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} - 1\begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} + 1\begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} det⁡(A)=2((3)(2)−(−1)(1))−1((1)(2)−(−1)(1))+1((1)(1)−(3)(1))\det(A) = 2((3)(2) - (-1)(1)) - 1((1)(2) - (-1)(1)) + 1((1)(1) - (3)(1)) det⁡(A)=2(6+1)−1(2+1)+1(1−3)\det(A) = 2(6 + 1) - 1(2 + 1) + 1(1 - 3) det⁡(A)=2(7)−1(3)+1(−2)\det(A) = 2(7) - 1(3) + 1(-2) det⁡(A)=14−3−2=9\det(A) = 14 - 3 - 2 = 9
  2. Trouver la matrice des cofacteurs.

    Les cofacteurs CijC_{ij} sont calculés comme Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}, où MijM_{ij} est le mineur.
    C11=+∣3−112∣=6−(−1)=7C_{11} = +\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} = 6 - (-1) = 7
    C12=−∣1−112∣=−(2−(−1))=−3C_{12} = -\begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} = -(2 - (-1)) = -3
    C13=+∣1311∣=1−3=−2C_{13} = +\begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = 1 - 3 = -2
    C21=−∣1112∣=−(2−1)=−1C_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = -(2 - 1) = -1
    C22=+∣2112∣=4−1=3C_{22} = +\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 4 - 1 = 3
    C23=−∣2111∣=−(2−1)=−1C_{23} = -\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = -(2 - 1) = -1
    C31=+∣113−1∣=−1−3=−4C_{31} = +\begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = -1 - 3 = -4
    $C_{

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