Produit $$A\times B$$ et $$B\times A$$ : matrice inversible - Exercice 4

$$\red{\text{Commençons par calculer :}}$$ $$A^{2} -5A+8I_{3}$$
$$A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {4} & {-6} & {-18} \\ {6} & {-11} & {-27} \\ {-4} & {8} & {16} \end{array}\right)-5\left(\begin{array}{ccc} {1} & {0} & {-3} \\ {1} & {-1} & {-6} \\ {-1} & {2} & {5} \end{array}\right)+8\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)$$
$$A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {4} & {-6} & {-18} \\ {6} & {-11} & {-27} \\ {-4} & {8} & {16} \end{array}\right)-\left(\begin{array}{ccc} {5} & {0} & {-15} \\ {5} & {-5} & {-30} \\ {-5} & {10} & {25} \end{array}\right)+\left(\begin{array}{ccc} {8} & {0} & {0} \\ {0} & {8} & {0} \\ {0} & {0} & {8} \end{array}\right)$$
$$A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right)$$
$$\red{\text{Calculons maintenant :}}$$ $$A\left(A^{2}-5A+8I_{3} \right)$$
$$A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {\red{1}} & {\red{0}} & {\red{-3}} \\ {\blue{1}} & {\blue{-1}} & {\blue{-6}} \\ {\green{-1}} & {\green{2}} & {\green{5}} \end{array}\right){\mathop{\left(\begin{array}{ccc} {\red{1}\times \pink{7}+\red{0}\times \pink{1}+\red{\left(-3\right)}\times \pink{1}} & {\red{1}\times \purple{\left(-6\right)}+\red{0}\times \purple{2}+\red{\left(-3\right)}\times \purple{\left(-2\right)}} & {\red{1}\times \orange{\left(-3\right)}+\red{0}\times \orange{3}+\red{\left(-3\right)}\times \orange{\left(-1\right)}} \\ {\blue{1}\times \pink{7}+\blue{\left(-1\right)}\times \pink{1}+\blue{\left(-6\right)}\times \pink{1}} & {\blue{1}\times \purple{\left(-6\right)}+\blue{\left(-1\right)}\times \purple{2}+\blue{\left(-6\right)}\times \purple{\left(-2\right)}} & {\blue{1}\times \orange{\left(-3\right)}+\blue{\left(-1\right)}\times \orange{3}+\blue{\left(-6\right)}\times \orange{\left(-1\right)}} \\ {\green{\left(-1\right)}\times \pink{7}+\green{2}\times \pink{1}+\green{5}\times \pink{1}} & {\green{\left(-1\right)}\times \purple{\left(-6\right)}+\green{2}\times \purple{2}+\green{5}\times \purple{\left(-2\right)}} & {\green{\left(-1\right)}\times \orange{\left(-3\right)}+\green{2}\times \orange{3}+\green{5}\times \orange{\left(-1\right)}} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {\pink{7}} & {\purple{-6}} & {\orange{-3}} \\ {\pink{1}} & {\purple{2}} & {\orange{3}} \\ {\pink{1}} & {\purple{-2}} & {\orange{-1}} \end{array}\right)}}$$
$$A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {1} & {0} & {-3} \\ {1} & {-1} & {-6} \\ {-1} & {2} & {5} \end{array}\right){\mathop{\left(\begin{array}{ccc} {4} & {0} & {0} \\ {0} & {4} & {0} \\ {0} & {0} & {4} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right)}}$$
$$A\left(A^{2} -5A+8I_{3} \right)=\left(\begin{array}{ccc} {4} & {0} & {0} \\ {0} & {4} & {0} \\ {0} & {0} & {4} \end{array}\right)$$
$$A\left(A^{2} -5A+8I_{3} \right)=4\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)$$
Finalement :

A l'aide de la calculatrice, on obtient :

D'après les questions $$1$$ et $$2$$, nous avons montré que : $$A\left(A^{2} -5A+8I_{3} \right)=4I_{3}$$ et $$\left(A^{2} -5A+8I_{3} \right)A=4I_{3}$$
$$\blue{\text{D'une part :}}$$
$$A\left(A^{2} -5A+8I_{3} \right)=4I_{3} \Leftrightarrow \frac{A\left(A^{2} -5A+8I_{3} \right)}{4} =I_{3} \Leftrightarrow A\times \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}=I_{3}$$
$$\blue{\text{D'autre part :}}$$
$$\left(A^{2} -5A+8I_{3} \right)A=4I_{3} \Leftrightarrow \frac{\left(A^{2} -5A+8I_{3} \right)A}{4} =I_{3} \Leftrightarrow \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}\times A=I_{3}$$
Nous avons donc $$A\times \pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}=I_{3}$$ $$\red{\text{et}}$$ $$\pink{\frac{1}{4} \left(A^{2} -5A+8I_{3} \right)}\times A=I_{3}$$
Il en résulte donc que $$A$$ est inversible et sa matrice inverse que l'on note $$A^{-1}$$ est égale à :
Comme $$A^{2} -5A+8I_{3} =\left(\begin{array}{ccc} {7} & {-6} & {-3} \\ {1} & {2} & {3} \\ {1} & {-2} & {-1} \end{array}\right)$$ alors