Matemática, perguntado por veronicapinheiro2001, 8 meses atrás

solução do sistema abaixo​

Anexos:

Soluções para a tarefa

Respondido por ShinyComet
2

    \left\{\begin{array}{III}{2y+3z=0}\\\\{x+4y-2z=0}\\\\{2x-3z=0}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{2y=-3z}\\\\{x=-4y+2z}\\\\{-3z=-2x}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-\dfrac{3z}{2}}\\\\{x=-4y+2z}\\\\{-3z=-2(-4y+2z)}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-\dfrac{3z}{2}}\\\\{x=-4y+2z}\\\\{-3z=8y-4z}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-\dfrac{3z}{2}}\\\\{x=-4y+2z}\\\\{-3z+4z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-\dfrac{3\times8y}{2}}\\\\{x=-4y+2z}\\\\{z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-\dfrac{3\times4y}{1}}\\\\{x=-4y+2z}\\\\{z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=-12y}\\\\{x=-4y+2z}\\\\{z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y+12y=0}\\\\{x=-4y+2z}\\\\{z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{13y=0}\\\\{x=-4y+2z}\\\\{z=8y}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=0}\\\\{x=-4y+2z}\\\\{z=8\times0}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=0}\\\\{x=-4\times0+2\times0}\\\\{z=0}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=0}\\\\{x=0+0}\\\\{z=0}\end{array}\Leftrightarrow

\Leftrightarrow\left\{\begin{array}{III}{y=0}\\\\{x=0}\\\\{z=0}\end{array}

Resposta:\;(x\;;\;y\;;\;z)=(0\;;\;0\;;\;0)

Podes ver mais resoluções de sistemas em:

  • https://brainly.com.br/tarefa/32072039
  • https://brainly.com.br/tarefa/31726917
  • https://brainly.com.br/tarefa/30133262
Anexos:
Perguntas interessantes