Exercices types : $$1$$^ère partie - Exercice 1

$$\red{\text{Commençons par calculer :}}$$ $$A^{2}-12A+44I_{3}$$
$$A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {26} & {10} & {-10} \\ {16} & {20} & {-16} \\ {6} & {-6} & {10} \end{array}\right)-12\left(\begin{array}{ccc} {5} & {1} & {-1} \\ {2} & {4} & {-2} \\ {1} & {-1} & {3} \end{array}\right)+44\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)$$
$$A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {26} & {10} & {-10} \\ {16} & {20} & {-16} \\ {6} & {-6} & {10} \end{array}\right)-\left(\begin{array}{ccc} {60} & {12} & {-12} \\ {24} & {48} & {-24} \\ {12} & {-12} & {36} \end{array}\right)+\left(\begin{array}{ccc} {44} & {0} & {0} \\ {0} & {44} & {0} \\ {0} & {0} & {44} \end{array}\right)$$
$$A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right)$$

$$\red{\text{Calculons maintenant :}}$$ $$A\left(A^{2} -12A+44I_{3} \right)$$
$$A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {\red{5}} & {\red{1}} & {\red{-1}} \\ {\blue{2}} & {\blue{4}} & {\blue{-2}} \\ {\green{1}} & {\green{-1}} & {\green{3}} \end{array}\right){\mathop{\left(\begin{array}{ccc} {\red{5}\times \pink{10}+\red{1}\times \pink{\left(-8\right)}+\red{\left(-1\right)}\times \pink{\left(-6\right)}} & {\red{5}\times \purple{\left(-2\right)}+\red{1}\times \purple{16}+\red{\left(-1\right)}\times \purple{6}} & {\red{5}\times \orange{2}+\red{1}\times \orange{8}+\red{\left(-1\right)}\times \orange{18}} \\ {\blue{2}\times \pink{10}+\blue{4}\times \pink{\left(-8\right)}+\blue{\left(-2\right)}\times \pink{\left(-6\right)}} & {\blue{2}\times \purple{\left(-2\right)}+\blue{4}\times \purple{16}+\blue{\left(-2\right)}\times \purple{6}} & {\blue{2}\times \orange{2}+\blue{4}\times \orange{8}+\blue{\left(-2\right)}\times \orange{18}} \\ {\green{1}\times \pink{10}+\green{\left(-1\right)}\times \pink{\left(-8\right)}+\green{3}\times \pink{\left(-6\right)}} & {\green{1}\times \purple{\left(-2\right)}+\green{\left(-1\right)}\times \purple{16}+\green{3}\times \purple{6}} & {\green{1}\times \orange{2}+\green{\left(-1\right)}\times \orange{8}+\green{3}\times \orange{18}} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {\pink{10}} & {\purple{-2}} & {\orange{2}} \\ {\pink{-8}} & {\purple{16}} & {\orange{8}} \\ {\pink{-6}} & {\purple{6}} & {\orange{18}} \end{array}\right)}}$$
$$A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {5} & {1} & {-1} \\ {2} & {4} & {-2} \\ {1} & {-1} & {3} \end{array}\right){\mathop{\left(\begin{array}{ccc} {48} & {0} & {0} \\ {0} & {48} & {0} \\ {0} & {0} & {48} \end{array}\right)}\limits^{\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right)}}$$
$$A\left(A^{2} -12A+44I_{3} \right)=\left(\begin{array}{ccc} {48} & {0} & {0} \\ {0} & {48} & {0} \\ {0} & {0} & {48} \end{array}\right)$$
$$A\left(A^{2} -12A+44I_{3} \right)=48\left(\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)$$

Finalement :

A l'aide de la calculatrice, on obtient :

D'après les questions $$1$$ et $$2$$, nous avons montré que : $$A\left(A^{2}-12A+44I_{3} \right)=48I_{3}$$ et $$\left(A^{2}-12A+44I_{3} \right)A=48I_{3}$$
$$\blue{\text{D'une part :}}$$
$$A\left(A^{2}-12A+44I_{3} \right)=48I_{3} \Leftrightarrow \frac{A\left(A^{2} -12A+44I_{3} \right)}{48} =I_{3} \Leftrightarrow A\times \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}=I_{3}$$
$$\blue{\text{D'autre part :}}$$
$$\left(A^{2} -12A+44I_{3} \right)A=48I_{3} \Leftrightarrow \frac{\left(A^{2} -12A+44I_{3} \right)A}{48} =I_{3} \Leftrightarrow \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}\times A=I_{3}$$
Nous avons donc $$A\times \pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}=I_{3}$$ $$\red{\text{et}}$$ $$\pink{\frac{1}{48} \left(A^{2} -12A+44I_{3} \right)}\times A=I_{3}$$
Il en résulte donc que $$A$$ est inversible et sa matrice inverse que l'on note $$A^{-1}$$ est égale à :
Comme $$A^{2} -12A+44I_{3} =\left(\begin{array}{ccc} {10} & {-2} & {2} \\ {-8} & {16} & {8} \\ {-6} & {6} & {18} \end{array}\right)$$ alors