Question

Determine os valores de x, y e z para solucionar op sistema linear a seguir:

$\left \{ {{x+y+z=0} \atop {3x+2y+3z=3}} \atop {x+3y+3z=1} \right.$

gabarito é

x=-1/2
y=-3
z=7/2

Answer 1

$\left\{ \!\begin{array}{ccl} x+y+z&\!\!=\!\!&0\\ 3x+2y+3z&\!\!=\!\!&3\\ x+3y+3z&\!\!=\!\!&1 \end{array} \right.$


Reescrevendo o sistema acima na forma matricial:

$\underbrace{\left[\begin{array}{ccc} 1&1&1\\ 3&2&3\\ 1&3&3 \end{array} \right ]}_{\mathbf{A}} \left[\begin{array}{c}x\\y\\z \end{array} \right ]=\left[\begin{array}{c}0\\3\\1 \end{array} \right ]$


• Calculando o determinante da matriz $\mathbf{A}$ dos coeficientes:

$\det \mathbf{A}=\det\left[\begin{array}{ccc} 1&1&1\\ 3&2&3\\ 1&3&3 \end{array} \right ]\\\\\\ \begin{array}{lr} \det\mathbf{A}=&1\cdot 2\cdot 3+1\cdot 3\cdot 1+1\cdot 3\cdot 3\\ &-1\cdot2\cdot 1-3\cdot 3\cdot 1-3\cdot 3\cdot 1 \end{array}\\\\\\ \det\mathbf{A}=6+3+9-2-9-9\\\\ \boxed{\begin{array}{c}\det\mathbf{A}=-2 \end{array}}\ne 0$


Como o determinante de $\mathbf{A}$ é diferente de zero, o sistema possui solução única (sistema possível e determinado - SPD).

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Agora, temos de calcular os determinantes correspondentes às variáveis $x,$ $y$ e $z.$

• Determinante correspondente à variável $x:$

Substituímos a 1ª coluna de $\mathbf{A}$ pelos termos independentes:

$D_x=\det\left[\begin{array}{ccc} 0&1&1\\ 3&2&3\\ 1&3&3 \end{array} \right ]\\\\\\ \begin{array}{lr} D_x=&0\cdot 2\cdot 3+1\cdot 3\cdot 1+1\cdot 3\cdot 3\\ &-1\cdot2\cdot 1-3\cdot 3\cdot 0-3\cdot 3\cdot 1 \end{array}\\\\\\ D_x=0+3+9-2-0-9\\\\ \boxed{\begin{array}{c}D_x=1 \end{array}}$


• Determinante correspondente à variável $y:$

Substituímos a 2ª coluna de $\mathbf{A}$ pelos termos independentes:

$D_y=\det\left[\begin{array}{ccc} 1&0&1\\ 3&3&3\\ 1&1&3 \end{array} \right ]\\\\\\ \begin{array}{lr} D_y=&1\cdot 3\cdot 3+0\cdot 3\cdot 1+1\cdot 3\cdot 1\\ &-1\cdot 3\cdot 1-1\cdot 3\cdot 1-3\cdot 3\cdot 0 \end{array}\\\\\\ D_y=9+0+3-3-3-0\\\\ \boxed{\begin{array}{c}D_y=6 \end{array}}$


• Determinante correspondente à variável $z:$

Substituímos a 3ª coluna de $\mathbf{A}$ pelos termos independentes:

$D_z=\det\left[\begin{array}{ccc} 1&1&0\\ 3&2&3\\ 1&3&1 \end{array} \right ]\\\\\\ \begin{array}{lr} D_z=&1\cdot 2\cdot 1+1\cdot 3\cdot 1+0\cdot 3\cdot 3\\ &-1\cdot 2\cdot 0-3\cdot 3\cdot 1-1\cdot 3\cdot 1 \end{array}\\\\\\ D_z=2+3+0-0-9-3\\\\ \boxed{\begin{array}{c}D_z=-7 \end{array}}$

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Sendo assim, pela Regra de Cramer, temos

$\bullet\;\;x=\dfrac{D_x}{\det\mathbf{A}}\\\\\\ x=\dfrac{1}{-2}\\\\\\ \boxed{\begin{array}{c}x=-\,\dfrac{1}{2}\end{array}}\\\\\\\\ \bullet\;\;y=\dfrac{D_y}{\det\mathbf{A}}\\\\\\ y=\dfrac{6}{-2}\\\\\\ \boxed{\begin{array}{c}y=-3\end{array}}\\\\\\\\ \bullet\;\;z=\dfrac{D_z}{\det\mathbf{A}}\\\\\\ z=\dfrac{-7}{-2}\\\\\\ \boxed{\begin{array}{c}z=\dfrac{7}{2}\end{array}}$


Bons estudos! :-)