Question

resolver e classificar o sistema linear :
x-4y+z=5
2×-y+3z=4
3×-2y-z=-2

Answer 1

Olá


$\left\{\begin{array}{lll}\mathsf{x-4y+z=5}~\\\mathsf{2x-y+3z=4}~\\\mathsf{3x-2y-z=-2}\end{array}\right$


Resolvendo pelo método do escalonamento


L2 = L2 - 2L1


$\left\{\begin{array}{lll}\mathsf{x-4y+z=5}~\\\mathsf{ \quad~~7y+z=-6}~\\\mathsf{3x-2y-z=-2}\end{array}\right$


L3 = L3 - 3L1


$\left\{\begin{array}{lll}\mathsf{x-4y+z=5}~\\\mathsf{ \quad~~7y+z=-6}~\\\mathsf{\quad ~~10y-4z=-17}\end{array}\right$



L3 = 7L3 - 10L2



$\displaystyle \left\{\begin{array}{lll}\mathsf{x-4y+z=5}~\\\mathsf{ \quad~~7y+z=-6}~\\\mathsf{\quad ~~-38z=-59}\end{array}\right\\\\\\\mathsf{-38z=-59}\qquad\Longrightarrow\qquad\boxed{\mathsf{z= \frac{59}{38} }}\\\\\\\\\text{da segunda equacao}\\\\\mathsf{7y+z=-6}\\\\\mathsf{7y+ \frac{59}{38}=-6 }\qquad\Longrightarrow\qquad\boxed{\mathsf{y=- \frac{41}{38} }}\\\\\\\\\text{da primeira equacao}\\\\\mathsf{x+4y+z=5}$


$\displaystyle \mathsf{x+4\left(- \frac{41}{38} \right)+ \frac{59}{38} =5}\qquad\Longrightarrow\qquad\boxed{\mathsf{x=- \frac{33}{38} }}\\\\\\\\\boxed{\mathsf{S=\left\{ \left(- \frac{33}{38} ,- \frac{41}{38},\frac{59}{38} \right)\right\}}}\\\\\\\text{Sistema Possivel e Determinado (SPD)}$