Question

Determinar a inversa de cada matriz abaixo na imagen:

Answer 1

Resposta:

De maneira geral, se temos uma matriz qualquer $X$,

$X=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)$

para encontrarmos a inversa $X^{-1}$ temos a seguinte regra,

$X^{-1}=\dfrac{1}{ad-bc}\left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)$

Onde $ad-bc$ é o determinante da matriz.

Vamos aos exercícios.

$A=\left(\begin{array}{ccc}5&6\\4&5\end{array}\right)$

$A^{-1}=\dfrac{1}{(5\cdot5)-(4\cdot6)}\left(\begin{array}{ccc}5&-6\\-4&5\end{array}\right)$

$A^{-1}=\dfrac{1}{25-24}\left(\begin{array}{ccc}5&-6\\-4&5\end{array}\right)$

$A^{-1}=\dfrac{1}{1}\left(\begin{array}{ccc}5&-6\\-4&5\end{array}\right)$

$A^{-1}=\left(\begin{array}{ccc}5&-6\\-4&5\end{array}\right)$

$B=\left(\begin{array}{ccc}2&5\\1&3\end{array}\right)$

$B^{-1}=\dfrac{1}{(2\cdot3)-(1\cdot5)}\left(\begin{array}{ccc}3&-5\\-1&2\end{array}\right)$

$B^{-1}=\dfrac{1}{6-5}\left(\begin{array}{ccc}3&-5\\-1&2\end{array}\right)$

$B^{-1}=\dfrac{1}{1}\left(\begin{array}{ccc}3&-5\\-1&2\end{array}\right)$

$B^{-1}=\left(\begin{array}{ccc}3&-5\\-1&2\end{array}\right)$

$C=\left(\begin{array}{ccc}1&0\\0&2\end{array}\right)$

$C^{-1}=\dfrac{1}{(1\cdot2)-(0\cdot0)}\left(\begin{array}{ccc}2&-0\\-0&1\end{array}\right)$

$C^{-1}=\dfrac{1}{2}\left(\begin{array}{ccc}2&0\\0&1\end{array}\right)$

$C^{-1}=\left(\begin{array}{ccc}\dfrac{1}{2}\cdot2&\dfrac{1}{2}\cdot0\\\dfrac{1}{2}\cdot0&\dfrac{1}{2}\cdot1\end{array}\right)$

$C^{-1}=\left(\begin{array}{ccc}\dfrac{2}{2}&0\\0&\dfrac{1}{2}\end{array}\right)$

$C^{-1}=\left(\begin{array}{ccc}1&0\\0&\dfrac{1}{2}\end{array}\right)$

$D=\left(\begin{array}{ccc}1&-1\\1&1\end{array}\right)$

$D^{-1}=\dfrac{1}{(1\cdot1)-[1\cdot(-1)]}\left(\begin{array}{ccc}1&-(-1)\\-1&1\end{array}\right)$

$D^{-1}=\dfrac{1}{1-(-1)}\left(\begin{array}{ccc}1&1\\-1&1\end{array}\right)$

$D^{-1}=\dfrac{1}{1+1}\left(\begin{array}{ccc}1&1\\-1&1\end{array}\right)$

$D^{-1}=\dfrac{1}{2}\left(\begin{array}{ccc}1&1\\-1&1\end{array}\right)$

$D^{-1}=\left(\begin{array}{ccc}\dfrac{1}{2}\cdot1&\dfrac{1}{2}\cdot1\\\dfrac{1}{2}\cdot(-1)&\dfrac{1}{2}\cdot1\end{array}\right)$

$D^{-1}=\left(\begin{array}{ccc}\dfrac{1}{2}&\dfrac{1}{2}\\-\dfrac{1}{2}&\dfrac{1}{2}\end{array}\right)$