Question

Resolva utilizando a regra de Cramer ou escalonamento, o sistema:

x+2y-z=1
2x+y+z=4
x-y+5z=5

Answer 1

Boa tarde

  x + 2y -    z = 1  (I)
2x +   y +   z = 4  (II)
  x -    y + 5z = 5  (III)

de (I) vem
z = x + 2y - 1
-z = -x - 2y + 1
5z = 5x + 10y - 5

de(II) vem
2x + y + x + 2y - 1 = 4
3x + 3y = 5 (IV)

de(III) vem
x - y + 5x + 10y - 5 = 5
6x + 9y = 10 (V)

de (IV) e (V) vem
3x + 3y = 5
6x + 9y = 10
6x + 6y = 10

y = 0

3x = 5
x = 5/3

z = x + 2y - 1 = 5/3 + 0 - 1 = 5/3 - 3/3  = 2/3 

S = (5/3, 0, 2/3)

Answer 2

Olá



Pelo método do escalonamento


$\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{2x+y+z=4}\\\mathsf{x-y+5z=5}\end{array}\right\\\\\\\mathsf{L2=L2-2L1}\\\\\\\text{Multiplica a primeira linha por -2 e soma com a segunda linha}\\\\\\\\\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~~-3y+3z=2}\\\mathsf{x-y+5z=5}\end{array}\right\\\\\\\\\text{L3=L2-L1}\\\\\text{Subtrai a linha 2 da linha 1}$

$\displaystyle\\\\\\\\\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~~-3y+3z=2}\\\mathsf{~~-3y+6z=4}\end{array}\right\\\\\\\text{L3=L3-L2}\\\\\\\text{Subtrai L3 de L2}\\\\\\\\\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~~-3y+z=2}\\\mathsf{~~~~~~~~~~~3z=2}\end{array}\right\\\\\\\\\text{Da 3}^a~\text{equacao}\\\\\\\mathsf{3z=2\qquad\qquad\Longrightarrow\qquad }\boxed{\mathsf{ z = \frac{2}{3} }}}$


Substituindo na 2ª equação


$\displaystyle\mathsf{-3y+3z=2}\\\\\\\mathsf{-3y+\diagup\!\!\!\!3\cdot \frac{2}{\diagup\!\!\!\!\!3} =2}\\\\\\\mathsf{-3y~=~2-2}\qquad\qquad\Longrightarrow\qquad\boxed{\mathsf{y=0}}$


Substituindo na 1ª equação


$\displaystyle\mathsf{x+2y-z=1}\\\\\\\mathsf{x+2\cdot 0 - \frac{2}{3}~=~1 }\\\\\\\mathsf{x = 1+ \frac{2}{3} }\qquad\qquad\Longrightarrow\qquad\boxed{\mathsf{x= \frac{5}{3} }}\\\\\\\\\\\text{Conjunto solucao}\\\\\\\boxed{\mathsf{S=\left\{ \frac{5}{3} ,~ 0, ~\frac{2}{3} \right\}}}$