Lista de exercícios do ensino médio para impressão
Resolver o sistema:$\phantom{X}\left\{\begin{array}{rcr} \;\dfrac{1}{x}\,+\,\dfrac{1}{y}\,-\,\dfrac{1}{z}\,=\,\phantom{X}4\;& \\ \;\dfrac{2}{x}\,-\,\dfrac{1}{y}\,+\,\dfrac{3}{z}\,=\,-3\;& \\ \; \dfrac{1}{x}\,+\,\dfrac{2}{y}\,+\,\dfrac{4}{z}\,=\,\phantom{X}1\;& \\ \end{array} \right.\,$

resposta:
Resolução:
Vamos fazer $\;p\,=\,\dfrac{1}{x}\;$, $\;q\,=\,\dfrac{1}{y}\;$ e $\;r\,=\,\dfrac{1}{z}\;$ e a seguir utilizar a Regra de Cramer.
Então temos o sistema:$\phantom{X}\left\{\begin{array}{rcr} \;p\;+\;q\;-\phantom{X}r\,=\phantom{X}4\;& \\ 2p\,-\;q\;+\,3r\,=\,-3\,& \\ \;p\;+\;2q\,+\,4r\,=\phantom{X}1\,& \\ \end{array} \right.\,$
● Calcular o valor do determinante D
$\;D\;=\,\begin{vmatrix} 1 & 1 & -1\; \\ 2 & -1 & 3 \; \\ 1 & 2 & 4 \;\end{vmatrix}\;=\;-20$
● Calcular o valor do determinante Dp
$\;D_p\;=\,\begin{vmatrix} 4 & 1 & -1\; \\ -3 & -1 & 3 \; \\ 1 & 2 & 4 \;\end{vmatrix}\;=\;-20$
● Calcular o valor do determinante Dq
$\;D_q\;=\,\begin{vmatrix} 1 & 4 & -1\; \\ 2 & -3 & 3 \; \\ 1 & 1 & 4 \;\end{vmatrix}\;=\;-40$
● Calcular o valor do determinante Dr
$\;D_r\;=\,\begin{vmatrix} 1 & 1 & 4\; \\ 2 & -1 & -3 \; \\ 1 & 2 & 1 \;\end{vmatrix}\;=\;20$
(calcular x)
$\;p\,=\,\dfrac{D_p}{D}\,=\,\dfrac{-20}{-20}\,=\,1\;\Rightarrow\;\boxed{\;x\,=\,\dfrac{1}{p}\,=\,1\;}$
(calcular y)
$\;q\,=\,\dfrac{D_q}{D}\,=\,\dfrac{-40}{-20}\,=\,2\;\Rightarrow\;\boxed{\;y\,=\,\dfrac{1}{q}\,=\,\dfrac{1}{2}\;}$
(calcular r)
$\;r\,=\,\dfrac{D_r}{D}\,=\,\dfrac{20}{-20}\,=\,-1\;\Rightarrow\;\boxed{\;z\,=\,\dfrac{1}{r}\,=\,-1\;}$
V = {(1, 1/2 , -1)}

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Veja exercÍcio sobre:
sistemas lineares
regra de Cramer