Question

Para que o sistema linear:

ax+y+z=3
x-2y+3z=2
x+2y-z=2

admita a solução é necessário que:

A) α ≠ 2

B) α ≠
$\frac{9}{4}$

C) α = 2

D) α =
$\frac{9}{4}$

E) ∀α∈R

Answer 1

Olá


Para facilitar nos cálculos, trocarei de posição a linha 1 pela linha 3


$\displaystyle \left\{\begin{array}{lll}\mathsf{x+2y-z=2}\\\mathsf{x-2y+3z=2}\\\mathsf{ax+y+z=3}\end{array}\right\\\\\\\mathsf{L2=L2-L1}\\\\\\\\\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~-4y+4z=0}\\\mathsf{ax+y+z=3}\end{array}\right\\\\\\\\\mathsf{L3=L3~-~aL1}\\\\\\\\\left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~-4y+4z=0}\\\mathsf{(1-2a)y+(1+a)z=3-2a}\end{array}\right\\\\\\\\\mathsf{L3=2L3~+~\left( \frac{1}{2}-a \right)L2}$

$\displaystyle \left\{\begin{array}{lll}\mathsf{x+2y-z=1}\\\mathsf{~-4y+4z=0}\\\mathsf{~~(4-2a)z=6-4a}\end{array}\right\\\\\\\mathsf{(4-2a)z = 6-4a}\\\\\\\mathsf{z= \frac{6-4a}{4-2a} }\\\\\\\mathsf{z= \frac{3-2a}{2-a} }$



Para que um sistema admita solução, o denominador tem que ser diferente de zero. Com isso, pode-se concluir que



α ≠ 2                             Letra A)