Question

1. resolva as esquações:
a) |x  2|    =   |6  3|
    |-1 x|         |2  x|

b) |2  x  2|   = |1  1  1|
    |3  x 0 |     |2   1  3|
    |5  1 1|      |4   1  9|

Answer 1

a) $\left|\begin{array}{ccc}x&2\\-1&x\\\end{array}\right|= \left|\begin{array}{ccc}6&3\\2&x\\\end{array}\right|$

Aplicando a regra para determinante de ordem 2, vem:

$x*x-(-1)*2=6*x-2*3$

$x^{2} +2=6x-6$

$x^{2} -6x+8=0$

Resolvendo esta equação do 2° grau, obtemos as raízes

$x'=2:::x''=4$

Logo:


$\boxed{\boxed{S=(2,~4)}}$ 



b) $\left|\begin{array}{ccc}2&x&2\\3&x&0\\5&1&1\end{array}\right|= \left|\begin{array}{ccc}1&1&1\\2&1&3\\4&1&9\end{array}\right|$

Aplicando a regra de Sarrus, em ambas as matrizes, vem:

$\left|\begin{array}{ccc}2&x&2\\3&x&0\\5&1&1\end{array}\right| \left\begin{array}{ccc}2&x\\3&x\\5&1\end{array}\right= \left|\begin{array}{ccc}1&1&1\\2&1&3\\4&1&9\end{array}\right| \left\begin{array}{ccc}1&1\\2&1\\4&1\end{array}\right$

$~\to~2x+0+6-10x-0-3x=9+12+2-4-3-18$

$~\to~-11x+6=-2$

$~\to~-11x=-8$

$\boxed{\boxed{x= \frac{8}{11}}}$


Espero ter ajudado e tenha bons estudos :)