Question

A matriz inversa de é igual a:

Answer 1

A matriz inversa será: $\large\displaystyle\begin{aligned} A^{-1}=\left[\begin{array}{ccc}6&4&-11\\5&3&-9\\-8&-5&14\end{array}\right] \end{aligned}$.

Para calcular a matriz $\bb A^{-1}$ , temos alguns métodos possíveis. O que eu irei utilizar será o da Matriz Adjunta. Esse método consiste que 4 passos. Sendo o primeiro calcular o determinante de A. Logo, dada a matriz:

$\large\displaystyle\begin{aligned} A=\left[\begin{array}{ccc}-3&-1&-3\\2&-4&-1\\-1&-2&-2\end{array}\right] \end{aligned}$

Seu determinante será:

$\large\displaystyle\text{ \begin{aligned} D(A)=\underbrace{\left|\begin{array}{ccc}-3&-1&-3\\2&-4&-1\\-1&-2&-2\end{array}\right| \left|\begin{array}{ccc}-3&-1\\2&-4\\-1&-2\end{array}\right|}_{\large\displaystyle\begin{aligned} D(A)= 1 \end{aligned}} \end{aligned} }$

• Agora, o segundo passo será calcular os cofatores da matriz A. Logo:

$\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-4&-1\\-2&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_1 = 6\end{aligned}} \end{aligned} }$  $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}2&-1\\-1&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_2 = -5\end{aligned}} \end{aligned} }$ $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}2&-4\\-1&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_3 = -8\end{aligned}} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-1&-3\\-2&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_4 = -4\end{aligned}} \end{aligned} }$ $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-3&-3\\-1&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_5 = 3\end{aligned}} \end{aligned} }$ $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-3&-1\\-1&-2\end{array}\right]}_{\large\displaystyle\begin{aligned} C_6 = 5\end{aligned}} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-1&-3\\-4&-1\end{array}\right]}_{\large\displaystyle\begin{aligned} C_7 = -11\end{aligned}} \end{aligned} }$ $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-3&-3\\2&-1\end{array}\right]}_{\large\displaystyle\begin{aligned} C_8 = 9\end{aligned}} \end{aligned} }$ $\large\displaystyle\text{ \begin{aligned}\underbrace{ \left[\begin{array}{ccc}-3&-1\\2&-4\end{array}\right]}_{\large\displaystyle\begin{aligned} C_9 = 14\end{aligned}} \end{aligned} }$

Feito isso, devemos agora criar uma matriz com os cofatores. Para criar a matriz dos cofatores, devemos seguir uma ordemzinha "( mantem e troca )" o sinal. Após isso calcular sua tramposta.

$\large\displaystyle\text{ \begin{aligned} \underbrace{\left[\begin{array}{ccc}6&5&-8\\4&3&-5\\-11&-9&14\end{array}\right]}_{\large\displaystyle\begin{aligned} Matriz\ dos\ cofatores.\end{aligned}} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \underbrace{\left[\begin{array}{ccc}6&4&-11\\5&3&-9\\-8&-5&14\end{array}\right]}_{\large\displaystyle\begin{aligned} Matriz\ transposta.\end{aligned}} \end{aligned} }$

Quarto passo. Pegue a matriz transposta dos cofatores e divida pelo determinante de A. Assim encontrando a matriz inversa. Logo:

$\large\displaystyle\text{ \begin{aligned} A^{-1}=\frac{\left[\begin{array}{ccc}6&4&-11\\5&3&-9\\-8&-5&14\end{array}\right]}{D(A)} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} A^{-1}=\frac{\left[\begin{array}{ccc}6&4&-11\\5&3&-9\\-8&-5&14\end{array}\right]}{1} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\green{A^{-1}=\left[\begin{array}{ccc}6&4&-11\\5&3&-9\\-8&-5&14\end{array}\right]}}} \end{aligned} }$

Veja mais sobre:

Matriz inversa.

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