Question

Meeeee Ajudeeem, é URGENTE trabalho tenho q entregar amanha cedo :( 
 
Se A-¹ a inversa de A= (7 -3)
                                  (2 -1)
a) A+A-¹
b) (A-¹)² + A²

Answer 1

$\boxed{\boxed{A^{-1}=(1/det~A)*adj~A}}$

$A= \left[\begin{array}{cc}7&-3\\2&-1\end{array}\right]\\\\det~A=7*(-1)-(-3)*2=-7+6=-1$

adj A = Matriz adjunta de A. Para descobrimos a adjunta dessa matriz, basta inverter a ordem das entradas da diagonal principal e trocar o sinal das entradas da diagonal secundária

$adj~A=\left[\begin{array}{cc}-1&3\\-2&7\end{array}\right]$

$A^{-1}=(1/det~A)*adj~A\\\\A^{-1}=(1/[-1])*\left[\begin{array}{cc}-1&3\\-2&7\end{array}\right]\\\\A^{-1}=(-1)*\left[\begin{array}{cc}-1&3\\-2&7\end{array}\right]\\\\A^{-1}=\left[\begin{array}{cc}1&-3\\2&-7\end{array}\right]$
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a)

$A+A^{-1}=\left[\begin{array}{cc}7&-3\\2&-1\end{array}\right] +\left[\begin{array}{cc}1&-3\\2&-7\end{array}\right]\\\\A+A^{-1}=\left[\begin{array}{cc}7+1&-3-3\\2+2&-1-7\end{array}\right]\\\\A+A^{-1}=\left[\begin{array}{cc}8&-6\\4&-8\end{array}\right]$

b)

$(A^{-1})^{2}=A^{-1}*A^{-1}\\\\(A^{-1})^{2}=\left[\begin{array}{cc}1&-3\\2&-7\end{array}\right]*\left[\begin{array}{cc}1&-3\\2&-7\end{array}\right]\\\\(A^{-1})^{2}=\left[\begin{array}{cc}1.1-3.2&1.(-3)-3.(-7)\\2.1-7.2&2.(-3)-7.(-7)\end{array}\right]\\\\(A^{-1})^{2}=\left[\begin{array}{cc}-5&18\\-12&43\end{array}\right]$

$A^{2}=A*A\\\\A^{2}=\left[\begin{array}{cc}7&-3\\2&-1\end{array}\right]*\left[\begin{array}{cc}7&-3\\2&-1\end{array}\right]\\\\A^{2}=\left[\begin{array}{cc}7.7-3.2&7.(-3)-3.(-1)\\2.7-1.2&2.(-3)-1.(-1)\end{array}\right]\\\\A^{2}=\left[\begin{array}{cc}43&-18\\12&-5\end{array}\right]$
_______________________

$(A^{-1})^{2} + A^{2}=\left[\begin{array}{cc}-5&18\\-12&43\end{array}\right]+\left[\begin{array}{cc}43&-18\\12&-5\end{array}\right]\\\\(A^{-1})^{2}+A^{2}=\left[\begin{array}{cc}-5+43&18-18\\-12+12&43-5\end{array}\right]\\\\(A^{-1})^{2}+A^{2}=\left[\begin{array}{cc}38&0\\0&38\end{array}\right]$