Bonjour,
Ex 8)
A₁⁻¹ =
Méthode : A x X = B ⇔ A⁻¹ x A x X = A⁻¹ x B ⇔ X = A⁻¹ x B
⇒ X = A₁⁻¹ x (1 0 1) (en colonne)
soit X₁ =
pour A₂ : A₂⁻¹ =
et X₂ =
Ex 9)
1) A inversible si det(A) ≠ 0
je note (a a a) la dernière ligne
det(A) = 1 * (a - a) - 3(a - a) + 2(a - a) = 0 donc non inversible
...bizarre vue la question 2 ??
2) A⁻¹ = com(A)/det(A) n'existe pas
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Bonjour,
Ex 8)
A₁⁻¹ =![\left[\begin{array}{ccc}-1&0&-1\\3/4&1/4&1/2\\1/2&-1/2&0\end{array}\right] \left[\begin{array}{ccc}-1&0&-1\\3/4&1/4&1/2\\1/2&-1/2&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%26-1%5C%5C3%2F4%261%2F4%261%2F2%5C%5C1%2F2%26-1%2F2%260%5Cend%7Barray%7D%5Cright%5D)
Méthode : A x X = B ⇔ A⁻¹ x A x X = A⁻¹ x B ⇔ X = A⁻¹ x B
⇒ X = A₁⁻¹ x (1 0 1) (en colonne)
soit X₁ =![\left[\begin{array}{ccc}-2\\5/4\\1/2\end{array}\right] \left[\begin{array}{ccc}-2\\5/4\\1/2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%5C%5C5%2F4%5C%5C1%2F2%5Cend%7Barray%7D%5Cright%5D)
pour A₂ : A₂⁻¹ =![\left[\begin{array}{ccc}4&-1&0\\-7&2&1\\10&-3&-1\end{array}\right] \left[\begin{array}{ccc}4&-1&0\\-7&2&1\\10&-3&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26-1%260%5C%5C-7%262%261%5C%5C10%26-3%26-1%5Cend%7Barray%7D%5Cright%5D)
et X₂ =![\left[\begin{array}{ccc}1\\-2\\1\end{array}\right] \left[\begin{array}{ccc}1\\-2\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C-2%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Ex 9)
1) A inversible si det(A) ≠ 0
je note (a a a) la dernière ligne
det(A) = 1 * (a - a) - 3(a - a) + 2(a - a) = 0 donc non inversible
...bizarre vue la question 2 ??
2) A⁻¹ = com(A)/det(A) n'existe pas