Exercices types : 11ère partie - Exercice 1

15 min
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COMPETENCES‾  :  1°)  Raisonner.{\color{red}\underline{COMPETENCES}\;:\;1°)\;Raisonner.}     \;\; 2°)  Calculer.{\color{red}2°)\;Calculer.}
Question 1
On considère la matrice A=(51−124−21−13)A=\left(\begin{array}{ccc} {5} & {1} & {-1} \\ {2} & {4} & {-2} \\ {1} & {-1} & {3} \end{array}\right) . On admet que A2=(2610−101620−166−610)A^{2} =\left(\begin{array}{ccc} {26} & {10} & {-10} \\ {16} & {20} & {-16} \\ {6} & {-6} & {10} \end{array}\right) .

Calculer A(A2−12A+44I3)A\left(A^{2}-12A+44I_{3} \right).

Correction
Commençons par calculer :\red{\text{Commençons par calculer :}} A2−12A+44I3A^{2}-12A+44I_{3}
A2−12A+44I3=(2610−101620−166−610)−12(51−124−21−13)+44(100010001)A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {26} & {10} & {-10} \\ {16} & {20} & {-16} \\ {6} & {-6} & {10} \end{array}\right)-12\left(\begin{array}{ccc} {5} & {1} & {-1} \\ {2} & {4} & {-2} \\ {1} & {-1} & {3} \end{array}\right)+44\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)
A2−12A+44I3=(2610−101620−166−610)−(6012−122448−2412−1236)+(440004400044)A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {26} & {10} & {-10} \\ {16} & {20} & {-16} \\ {6} & {-6} & {10} \end{array}\right)-\left(\begin{array}{ccc} {60} & {12} & {-12} \\ {24} & {48} & {-24} \\ {12} & {-12} & {36} \end{array}\right)+\left(\begin{array}{ccc} {44} & {0} & {0} \\ {0} & {44} & {0} \\ {0} & {0} & {44} \end{array}\right)
A2−12A+44I3=(10−22−8168−6618)A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right)

Calculons maintenant :\red{\text{Calculons maintenant :}} A(A2−12A+44I3)A\left(A^{2} -12A+44I_{3} \right)
A(A2−12A+44I3)=(51−124−21−13)(5×10+1×(−8)+(−1)×(−6)5×(−2)+1×16+(−1)×65×2+1×8+(−1)×182×10+4×(−8)+(−2)×(−6)2×(−2)+4×16+(−2)×62×2+4×8+(−2)×181×10+(−1)×(−8)+3×(−6)1×(−2)+(−1)×16+3×61×2+(−1)×8+3×18)(10−22−8168−6618)A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {\red{5}} & {\red{1}} & {\red{-1}} \\ {\blue{2}} & {\blue{4}} & {\blue{-2}} \\ {\green{1}} & {\green{-1}} & {\green{3}} \end{array}\right){\mathop{\left(\begin{array}{ccc} {\red{5}\times \pink{10}+\red{1}\times \pink{\left(-8\right)}+\red{\left(-1\right)}\times \pink{\left(-6\right)}} & {\red{5}\times \purple{\left(-2\right)}+\red{1}\times \purple{16}+\red{\left(-1\right)}\times \purple{6}} & {\red{5}\times \orange{2}+\red{1}\times \orange{8}+\red{\left(-1\right)}\times \orange{18}} \\ {\blue{2}\times \pink{10}+\blue{4}\times \pink{\left(-8\right)}+\blue{\left(-2\right)}\times \pink{\left(-6\right)}} & {\blue{2}\times \purple{\left(-2\right)}+\blue{4}\times \purple{16}+\blue{\left(-2\right)}\times \purple{6}} & {\blue{2}\times \orange{2}+\blue{4}\times \orange{8}+\blue{\left(-2\right)}\times \orange{18}} \\ {\green{1}\times \pink{10}+\green{\left(-1\right)}\times \pink{\left(-8\right)}+\green{3}\times \pink{\left(-6\right)}} & {\green{1}\times \purple{\left(-2\right)}+\green{\left(-1\right)}\times \purple{16}+\green{3}\times \purple{6}} & {\green{1}\times \orange{2}+\green{\left(-1\right)}\times \orange{8}+\green{3}\times \orange{18}} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {\pink{10}} & {\purple{-2}} & {\orange{2}} \\ {\pink{-8}} & {\purple{16}} & {\orange{8}} \\ {\pink{-6}} & {\purple{6}} & {\orange{18}} \end{array}\right)}}
A(A2−12A+44I3)=(51−124−21−13)(480004800048)(10−22−8168−6618)A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {5} & {1} & {-1} \\ {2} & {4} & {-2} \\ {1} & {-1} & {3} \end{array}\right){\mathop{\left(\begin{array}{ccc} {48} & {0} & {0} \\ {0} & {48} & {0} \\ {0} & {0} & {48} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right)}}
A(A2−12A+44I3)=(480004800048)A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {48} & {0} & {0} \\ {0} & {48} & {0} \\ {0} & {0} & {48} \end{array}\right)
A(A2−12A+44I3)=48(100010001)A\left(A^{2} -12A+44I_{3} \right)=48\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)

Finalement :
A(A2−12A+44I3)=48I3A\left(A^{2} -12A+44I_{3} \right)=48I_{3}

Question 2

A l'aide de la calculatrice, calculer (A2−12A+44I3)A\left(A^{2}-12A+44I_{3} \right)A

Correction
A l'aide de la calculatrice, on obtient :
(A2−12A+44I3)A=48I3\left(A^{2}-12A+44I_{3} \right)A=48I_{3}
Question 3

En déduire que la matrice AA est inversible et déterminer alors son inverse.

Correction
  • Une matrice carrée d'ordre AA est inversible si et seulement si il existe une matrice BB telle que A×B=InA\times B=I_{n} et\red{\text{et}} B×A=InB\times A=I_{n} où InI_{n} correspond à la matrice identité .
    • D'après les questions 11 et 22, nous avons montré que : A(A2−12A+44I3)=48I3A\left(A^{2}-12A+44I_{3} \right)=48I_{3} et (A2−12A+44I3)A=48I3\left(A^{2}-12A+44I_{3} \right)A=48I_{3}
    D’une part :\blue{\text{D'une part :}}
    A(A2−12A+44I3)=48I3⇔A(A2−12A+44I3)48=I3⇔A×148(A2−12A+44I3)=I3A\left(A^{2}-12A+44I_{3} \right)=48I_{3} \Leftrightarrow \frac{A\left(A^{2} -12A+44I_{3} \right)}{48} =I_{3} \Leftrightarrow A\times \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}=I_{3}
    D’autre part :\blue{\text{D'autre part :}}
    (A2−12A+44I3)A=48I3⇔(A2−12A+44I3)A48=I3⇔148(A2−12A+44I3)×A=I3\left(A^{2} -12A+44I_{3} \right)A=48I_{3} \Leftrightarrow \frac{\left(A^{2} -12A+44I_{3} \right)A}{48} =I_{3} \Leftrightarrow \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}\times A=I_{3}
    Nous avons donc A×148(A2−12A+44I3)=I3A\times \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}=I_{3} et\red{\text{et}} 148(A2−12A+44I3)×A=I3\pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}\times A=I_{3}
    Il en résulte donc que AA est inversible et sa matrice inverse que l'on note A−1A^{-1} est égale à :
    A−1=148(A2−12A+44I3)A^{-1}=\pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}

    Comme A2−12A+44I3=(10−22−8168−6618)A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right) alors
    A−1=(524−124124−161316−181838)A^{-1} =\left(\begin{array}{ccc} {\frac{5}{24} } & {-\frac{1}{24} } & {\frac{1}{24} } \\ {-\frac{1}{6} } & {\frac{1}{3} } & {\frac{1}{6} } \\ {-\frac{1}{8} } & {\frac{1}{8} } & {\frac{3}{8} } \end{array}\right)

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