Question

solução do sistema abaixo​

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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