Question

Verifique se existe $A^{-1}$ e , em caso afirmativo, calcule det $A^{-1}$.

ALTERNATIVA A- A= $\left[\begin{array}{ccc}1&3\\5&4\\\end{array}\right]$ det $A^{-1}$ = ___________

ALTERNATIVA B A= $\left[\begin{array}{ccc}-2&3\\-6&-9\\\end{array}\right]$

se responder qualquer coisa vou denunciar

Answer 1

Resposta:

Explicação passo-a-passo:

A)

$\left(\begin{matrix}1 & 3 \\5 & 4\end{matrix}\right)^{-1} = a(1,1)*a(2,2)-a (1,2)*a(2,1)$

$\left(\begin{matrix}1 & 3 \\5 & 4\end{matrix}\right) = \dfrac{1}{1*4 - 3*5} * \left(\begin{matrix}4 & -3 \\-5 & 1\end{matrix}\right) ~~=> ~~\left(\begin{matrix}\dfrac{-4}{11} & \dfrac{3}{11} \\\\\dfrac{5}{11} & \dfrac{-1}{11}\end{matrix}\right)$

$\left|\begin{matrix}\dfrac{-4}{11} & \dfrac{3}{11} \\\\ \dfrac{5}{11} & \dfrac{-1}{11}\end{matrix}\right| => \dfrac{-4}{11}*(\dfrac{-1}{11} )- \dfrac{3}{11} * \dfrac{5}{11} = \dfrac{-1}{11}$

===

B)

$\left(\begin{matrix}-2 & 3 \\-6 & -9\end{matrix}\right)^{-1} = \dfrac{1}{-2 * (-9) - (-3) * (-6) } = \left(\begin{matrix}-9 & -3 \\6 & -2\end{matrix}\right) ~~=>~~\left(\begin{matrix}\dfrac{-1}{4} & \dfrac{-1}{12} \\\\\dfrac{1}{6} & \dfrac{-1}{18}\end{matrix}\right)$

$\left(\begin{matrix}\dfrac{-1}{4} & \dfrac{-1}{12} \\\\\dfrac{1}{6} & \dfrac{-1}{18}\end{matrix}\right) = \dfrac{1}{4} * \dfrac{-1}{18} -\dfrac{-1}{12} * \dfrac{1}{6} ~~ = \dfrac{1}{36}$