Question

A equação | 2     1     3 |
                | 4     -1  n-1 |   = 12 tem como conjunto solução: 
                | n     0     n |      

a) { -6,2}
b) { -2,6}
c) {2,6}
d) {6,-6}
e) {-2,2}

ME AJUDEM POR FAVOR

Answer 1

Olá Louise,

use a mesma regra utilizada para determinante de 3ª ordem (regra de Sarruz):

$~~~~\searrow~~\searrow~~\searrow~~~\swarrow~~~\swarrow~~~\swarrow \\ ~~~~~~\left|\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right| \left\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right\\ ~~~~~\swarrow~~~\swarrow~~~\swarrow~\searrow~~~\searrow~~~\searrow\\\\ ~~\boxed{-}~~~\boxed{-}~~~\boxed{-}~~~~~~~~\boxed{+}~~~~\boxed{+}~~~\boxed{+}$

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$\left|\begin{array}{ccc}2&~~1&3\\4&-1&n-1\\n&~~0&n\end{array}\right|\left\begin{array}{ccc}2&~~1\\4&-1\\n&~~0\end{array}\right =12\\\\\\ -2n+n^2-n+0+3n-0-4n=12\\ n^2-2n-n+3n-4n=12\\ n^2-4n-12=0\\\\ \Delta=(-4)^2-4*1*(-12)\\ \Delta=16+48\\ \Delta=64\\\\ n= \dfrac{-(-4)\pm \sqrt{64} }{2*1}= \dfrac{4\pm8}{2}\begin{cases} n'=\dfrac{4-8}{2}= \dfrac{-4}{~~2} = -2\\\\ n''= \dfrac{4+8}{2}= \dfrac{12}{2}=6 \end{cases}$

E portanto, alternativa B,

$\boxed{S=\{-2,~6\}}$

Tenha ótimos estudos =))