Question

3.Utilizando a Regra de Cramer, resolva os seguintes sistemas:

a) { 3x - 4y = 1
{ 2x - 4y = 5

Answer 1

Método de Cramer:

3x - 4y = 1
2x - 4y = 5

D = - 4

[3  - 4]   =  3.(-4) - [ (- 4).2] = - 12 - [- 8] = - 12 + 8 = - 4
[2  - 4]
-------------------------------------

Dx =  16

Dx = 
[1   - 4]   =   1.(-4) - [(-4).5] = - 4 - [ -20] = - 4 + 20 = 16
[5   - 4]
------------------------------------------------------------------
Dy = 13

[3   1]   = 3.5 - 1.2 = 15 - 2 = 13
[2   5]
-----------------------------------------------------------

x = Dx/D
x = 16/- 4
x = - 4

y = Dy/D
y = 13/-4
y = - 13/4 
======================================

Método da Adição:

3x - 4y = 1
2x - 4y = 5  ( - 1)

 3x - 4y = 1
- 2x + 4y = - 5
x = - 4

3x - 4y = 1
3.(-4) - 4y = 1
- 12 - 4y = 1
- 12 - 1 = 4y
- 13 = 4y
4y = - 13
y = - 13/4

Answer 2

$\begin{cases}3x - 4y = 1\\ 2x - 4y = 5\end{cases}\\\\\\\\ D= \left[\begin{array}{ccc}3&-4\\2&-4\end{array}\right] \to~~\\\\\\D= [3\times(-4)]-[2\times (-4)]\to \\D=(-12)-(-8)\to\\ D=-12+8\to\\ \boxed{D=-4}$



$D_x= \left[\begin{array}{ccc}1&-4\\5&-4\end{array}\right] \to~~\\\\\\D_x= [1\times(-4)]-[5\times (-4)]\to \\D_x=(-4)-(-20)\to\\ D_x=-4+20\to\\ \boxed{D_x=16}$



$D_y= \left[\begin{array}{ccc}3&1\\2&5\end{array}\right] \to~~\\\\\\D_y= [3\times5]-[2\times1]\to \\D_y=15-2\to\\ \boxed{D_y=13}$




$x= \dfrac{D_x}{D}\to~~ x= \dfrac{16}{-4}\to~~ \boxed{x=-4}\\\\\\ y= \dfrac{D_y}{D}\to~~ y= \dfrac{13}{-4}\to~~ \boxed{y=-\dfrac{13}{4}}\\\\\\\\ \large\boxed{\boxed{S=\left\{-4~;~\dfrac{13}{4} \right\}}}$