Calcul matriciel

Produit A×BA\times B et B×AB\times A : matrice inversible - Exercice 4

15 min
30
Question 1
On considère la matrice A=(10−31−1−6−125)A=\left(\begin{array}{ccc} {1} & {0} & {-3} \\ {1} & {-1} & {-6} \\ {-1} & {2} & {5} \end{array}\right) . On admet que A2=(4−6−186−11−27−4816)A^{2} =\left(\begin{array}{ccc} {4} & {-6} & {-18} \\ {6} & {-11} & {-27} \\ {-4} & {8} & {16} \end{array}\right)

Calculer A(A2−5A+8I3)A\left(A^{2}-5A+8I_{3} \right).

Correction
Commençons par calculer :\red{\text{Commençons par calculer :}} A2−5A+8I3A^{2} -5A+8I_{3}
A2−5A+8I3=(4−6−186−11−27−4816)−5(10−31−1−6−125)+8(100010001)A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {4} & {-6} & {-18} \\ {6} & {-11} & {-27} \\ {-4} & {8} & {16} \end{array}\right)-5\left(\begin{array}{ccc} {1} & {0} & {-3} \\ {1} & {-1} & {-6} \\ {-1} & {2} & {5} \end{array}\right)+8\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)
A2−5A+8I3=(4−6−186−11−27−4816)−(50−155−5−30−51025)+(800080008)A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {4} & {-6} & {-18} \\ {6} & {-11} & {-27} \\ {-4} & {8} & {16} \end{array}\right)-\left(\begin{array}{ccc} {5} & {0} & {-15} \\ {5} & {-5} & {-30} \\ {-5} & {10} & {25} \end{array}\right)+\left(\begin{array}{ccc} {8} & {0} & {0} \\ {0} & {8} & {0} \\ {0} & {0} & {8} \end{array}\right)
A2−5A+8I3=(7−6−31231−2−1)A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right)
Calculons maintenant :\red{\text{Calculons maintenant :}} A(A2−5A+8I3)A\left(A^{2}-5A+8I_{3} \right)
A(A2−5A+8I3)=(10−31−1−6−125)(1×7+0×1+(−3)×11×(−6)+0×2+(−3)×(−2)1×(−3)+0×3+(−3)×(−1)1×7+(−1)×1+(−6)×11×(−6)+(−1)×2+(−6)×(−2)1×(−3)+(−1)×3+(−6)×(−1)(−1)×7+2×1+5×1(−1)×(−6)+2×2+5×(−2)(−1)×(−3)+2×3+5×(−1))(7−6−31231−2−1)A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {\red{1}} & {\red{0}} & {\red{-3}} \\ {\blue{1}} & {\blue{-1}} & {\blue{-6}} \\ {\green{-1}} & {\green{2}} & {\green{5}} \end{array}\right){\mathop{\left(\begin{array}{ccc} {\red{1}\times \pink{7}+\red{0}\times \pink{1}+\red{\left(-3\right)}\times \pink{1}} & {\red{1}\times \purple{\left(-6\right)}+\red{0}\times \purple{2}+\red{\left(-3\right)}\times \purple{\left(-2\right)}} & {\red{1}\times \orange{\left(-3\right)}+\red{0}\times \orange{3}+\red{\left(-3\right)}\times \orange{\left(-1\right)}} \\ {\blue{1}\times \pink{7}+\blue{\left(-1\right)}\times \pink{1}+\blue{\left(-6\right)}\times \pink{1}} & {\blue{1}\times \purple{\left(-6\right)}+\blue{\left(-1\right)}\times \purple{2}+\blue{\left(-6\right)}\times \purple{\left(-2\right)}} & {\blue{1}\times \orange{\left(-3\right)}+\blue{\left(-1\right)}\times \orange{3}+\blue{\left(-6\right)}\times \orange{\left(-1\right)}} \\ {\green{\left(-1\right)}\times \pink{7}+\green{2}\times \pink{1}+\green{5}\times \pink{1}} & {\green{\left(-1\right)}\times \purple{\left(-6\right)}+\green{2}\times \purple{2}+\green{5}\times \purple{\left(-2\right)}} & {\green{\left(-1\right)}\times \orange{\left(-3\right)}+\green{2}\times \orange{3}+\green{5}\times \orange{\left(-1\right)}} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {\pink{7}} & {\purple{-6}} & {\orange{-3}} \\ {\pink{1}} & {\purple{2}} & {\orange{3}} \\ {\pink{1}} & {\purple{-2}} & {\orange{-1}} \end{array}\right)}}
A(A2−5A+8I3)=(10−31−1−6−125)(400040004)(7−6−31231−2−1)A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {1} & {0} & {-3} \\ {1} & {-1} & {-6} \\ {-1} & {2} & {5} \end{array}\right){\mathop{\left(\begin{array}{ccc} {4} & {0} & {0} \\ {0} & {4} & {0} \\ {0} & {0} & {4} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right)}}
A(A2−5A+8I3)=(400040004)A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {4} & {0} & {0} \\ {0} & {4} & {0} \\ {0} & {0} & {4} \end{array}\right)
A(A2−5A+8I3)=4(100010001)A\left(A^{2} -5A+8I_{3} \right)=4\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)
Finalement :
A(A2−5A+8I3)=4I3A\left(A^{2} -5A+8I_{3} \right)=4I_{3}

Question 2

A l'aide de la calculatrice, calculer (A2−5A+8I3)A\left(A^{2}-5A+8I_{3} \right)A

Correction
A l'aide de la calculatrice, on obtient :
(A2−5A+8I3)A=4I3\left(A^{2} -5A+8I_{3} \right)A=4I_{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(A2−5A+8I3)=4I3A\left(A^{2} -5A+8I_{3} \right)=4I_{3} et (A2−5A+8I3)A=4I3\left(A^{2} -5A+8I_{3} \right)A=4I_{3}
    D’une part :\blue{\text{D'une part :}}
    A(A2−5A+8I3)=4I3⇔A(A2−5A+8I3)4=I3⇔A×14(A2−5A+8I3)=I3A\left(A^{2} -5A+8I_{3} \right)=4I_{3} \Leftrightarrow \frac{A\left(A^{2} -5A+8I_{3} \right)}{4} =I_{3} \Leftrightarrow A\times \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}=I_{3}
    D’autre part :\blue{\text{D'autre part :}}
    (A2−5A+8I3)A=4I3⇔(A2−5A+8I3)A4=I3⇔14(A2−5A+8I3)×A=I3\left(A^{2} -5A+8I_{3} \right)A=4I_{3} \Leftrightarrow \frac{\left(A^{2} -5A+8I_{3} \right)A}{4} =I_{3} \Leftrightarrow \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}\times A=I_{3}
    Nous avons donc A×14(A2−5A+8I3)=I3A\times \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}=I_{3} et\red{\text{et}} 14(A2−5A+8I3)×A=I3\pink{\frac{1}{4} \left(A^{2} -5A+8I_{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=14(A2−5A+8I3)A^{-1}=\pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}

    Comme A2−5A+8I3=(7−6−31231−2−1)A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right) alors
    A−1=(74−32−3414123414−12−14)A^{-1} =\left(\begin{array}{ccc} {\frac{7}{4} } & {-\frac{3}{2} } & {-\frac{3}{4} } \\ {\frac{1}{4} } & {\frac{1}{2} } & {\frac{3}{4} } \\ {\frac{1}{4} } & {-\frac{1}{2} } & {-\frac{1}{4} } \end{array}\right)