Question

Valendo 30 pontos!!
Determinante em R:

a) $\left[\begin{array}{ccc}x&1\\4&3\\\end{array}\right] = \left[\begin{array}{ccc}2x&1&\\\\3&4\end{array}\right]$

b) $\left[\begin{array}{ccc}2^{x+1}&2^x\\4&3\\\end{array}\right] = 16$

Answer 1

Olá,

use o mesmo procedimento utilizado para resolução de determinante de ordem 2:

$\left|\begin{array}{ccc}x&1\\4&3\\\end{array}\right|= \left|\begin{array}{ccc}2x&1\\3&4\\\end{array}\right|\\\\\\ 3\cdot x-4\cdot1=8\cdot x-3\cdot1\\ 3x-4=8x-3\\ 3x-8x=-3+4\\ -5x=1\\\\ x= -\dfrac{1}{5}\\\\ \large\boxed{\boxed{S=\left\{ -\dfrac{1}{5}\right\}}}$

>>>>>>>>>>>>>>>>>>>>>>>>

Essa equação necessita de aplicações de exponenciais, veja as propriedades que serão utilizadas:

$.............................................\\a^{m+n}~\to~a^m\cdot a^n\\\\ b\cdot a^m+c\cdot a^m~\to~a^m(b+c).\\ .............................................$

$\left|\begin{array}{ccc}2^{x+1}&2^x\\4&3\\\end{array}\right|=16\\\\\\ 3\cdot2^{x+1}-4\cdot2^x=16\\ 3\cdot2^x\cdot2^1-4\cdot2^x=16~~~~~.\\ 3\cdot2\cdot2^x-4\cdot2^x=16\\ 6\cdot2^x-4\cdot2^x=16\\ 2^x\cdot(6-4)=16\\ 2^x\cdot2=16\\\\ 2^x= \dfrac{16}{2}\\\\ 2^x=8\\ \not2^x=\not2^3\\\\ x=3\\\\\\ \huge\boxed{\boxed{S=\{3\}}}$

Tenha ótimos estudos =))