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Solución de sistemas de ecuaciones/ 2x+y-3z+4q=1; 3x+2y+4z-3q=-1; x-3y-z-2q=0; x+15y+5z+9q=0

2x+y-3z+4q=1; 3x+2y+4z-3q=-1; x-3y-z-2q=0; x+15y+5z+9q=0

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Gráfico:

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interior superior

Solución

Ha introducido [src] 
2*x + y - 3*z + 4*q = 1
4q+(−3z+(2x+y))=14 q + \left(- 3 z + \left(2 x + y\right)\right) = 1 
3*x + 2*y + 4*z - 3*q = -1
−3q+(4z+(3x+2y))=−1- 3 q + \left(4 z + \left(3 x + 2 y\right)\right) = -1 
x - 3*y - z - 2*q = 0
−2q+(−z+(x−3y))=0- 2 q + \left(- z + \left(x - 3 y\right)\right) = 0 
x + 15*y + 5*z + 9*q = 0
9q+(5z+(x+15y))=09 q + \left(5 z + \left(x + 15 y\right)\right) = 0 
9*q + 5*z + x + 15*y = 0
Respuesta rápida
y1=−59z67−1167y_{1} = - \frac{59 z}{67} - \frac{11}{67}
=
−59z67−1167- \frac{59 z}{67} - \frac{11}{67}
=
-0.164179104477612 - 0.880597014925373*z

x1=10z67+367x_{1} = \frac{10 z}{67} + \frac{3}{67}
=
10z67+367\frac{10 z}{67} + \frac{3}{67}
=
0.0447761194029851 + 0.149253731343284*z

q1=60z67+1867q_{1} = \frac{60 z}{67} + \frac{18}{67}
=
60z67+1867\frac{60 z}{67} + \frac{18}{67}
=
0.26865671641791 + 0.895522388059702*z
Método de Gauss
Tenemos el sistema de ecuaciones
4q+(−3z+(2x+y))=14 q + \left(- 3 z + \left(2 x + y\right)\right) = 1
−3q+(4z+(3x+2y))=−1- 3 q + \left(4 z + \left(3 x + 2 y\right)\right) = -1
−2q+(−z+(x−3y))=0- 2 q + \left(- z + \left(x - 3 y\right)\right) = 0
9q+(5z+(x+15y))=09 q + \left(5 z + \left(x + 15 y\right)\right) = 0

Expresamos el sistema de ecuaciones en su forma canónica
4q+2x+y−3z=14 q + 2 x + y - 3 z = 1
−3q+3x+2y+4z=−1- 3 q + 3 x + 2 y + 4 z = -1
−2q+x−3y−z=0- 2 q + x - 3 y - z = 0
9q+x+15y+5z=09 q + x + 15 y + 5 z = 0
Presentamos el sistema de ecuaciones lineales como matriz
[421−31−3324−1−21−3−10911550]\left[\begin{matrix}4 & 2 & 1 & -3 & 1\\-3 & 3 & 2 & 4 & -1\\-2 & 1 & -3 & -1 & 0\\9 & 1 & 15 & 5 & 0\end{matrix}\right]
En 1 de columna
[4−3−29]\left[\begin{matrix}4\\-3\\-2\\9\end{matrix}\right]
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
[421−31]\left[\begin{matrix}4 & 2 & 1 & -3 & 1\end{matrix}\right]
,
y lo restaremos de otras filas:
De 2 de fila restamos:
[−3−(−3)443−(−3)242−−344−−−94−1−−34]=[09211474−14]\left[\begin{matrix}-3 - \frac{\left(-3\right) 4}{4} & 3 - \frac{\left(-3\right) 2}{4} & 2 - - \frac{3}{4} & 4 - - \frac{-9}{4} & -1 - - \frac{3}{4}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\end{matrix}\right]
obtenemos
[421−3109211474−14−21−3−10911550]\left[\begin{matrix}4 & 2 & 1 & -3 & 1\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\-2 & 1 & -3 & -1 & 0\\9 & 1 & 15 & 5 & 0\end{matrix}\right]
De 3 de fila restamos:
[−2−(−1)421−(−1)22−3−−12−(−1)(−1)32−1−−12]=[02−52−5212]\left[\begin{matrix}-2 - \frac{\left(-1\right) 4}{2} & 1 - \frac{\left(-1\right) 2}{2} & -3 - - \frac{1}{2} & - \frac{\left(-1\right) \left(-1\right) 3}{2} - 1 & - \frac{-1}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 2 & - \frac{5}{2} & - \frac{5}{2} & \frac{1}{2}\end{matrix}\right]
obtenemos
[421−3109211474−1402−52−5212911550]\left[\begin{matrix}4 & 2 & 1 & -3 & 1\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 2 & - \frac{5}{2} & - \frac{5}{2} & \frac{1}{2}\\9 & 1 & 15 & 5 & 0\end{matrix}\right]
De 4 de fila restamos:
[9−4⋅941−2⋅94(−1)94+155−−274(−1)94]=[0−72514474−94]\left[\begin{matrix}9 - \frac{4 \cdot 9}{4} & 1 - \frac{2 \cdot 9}{4} & \frac{\left(-1\right) 9}{4} + 15 & 5 - - \frac{27}{4} & \frac{\left(-1\right) 9}{4}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{7}{2} & \frac{51}{4} & \frac{47}{4} & - \frac{9}{4}\end{matrix}\right]
obtenemos
[421−3109211474−1402−52−52120−72514474−94]\left[\begin{matrix}4 & 2 & 1 & -3 & 1\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 2 & - \frac{5}{2} & - \frac{5}{2} & \frac{1}{2}\\0 & - \frac{7}{2} & \frac{51}{4} & \frac{47}{4} & - \frac{9}{4}\end{matrix}\right]
En 2 de columna
[2922−72]\left[\begin{matrix}2\\\frac{9}{2}\\2\\- \frac{7}{2}\end{matrix}\right]
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
[09211474−14]\left[\begin{matrix}0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[4−0⋅492−4⋅92⋅91−4⋅114⋅9−3−4⋅74⋅91−−19]=[40−29−349109]\left[\begin{matrix}4 - \frac{0 \cdot 4}{9} & 2 - \frac{4 \cdot 9}{2 \cdot 9} & 1 - \frac{4 \cdot 11}{4 \cdot 9} & -3 - \frac{4 \cdot 7}{4 \cdot 9} & 1 - - \frac{1}{9}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & - \frac{2}{9} & - \frac{34}{9} & \frac{10}{9}\end{matrix}\right]
obtenemos
[40−29−34910909211474−1402−52−52120−72514474−94]\left[\begin{matrix}4 & 0 & - \frac{2}{9} & - \frac{34}{9} & \frac{10}{9}\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 2 & - \frac{5}{2} & - \frac{5}{2} & \frac{1}{2}\\0 & - \frac{7}{2} & \frac{51}{4} & \frac{47}{4} & - \frac{9}{4}\end{matrix}\right]
De 3 de fila restamos:
[−0⋅492−4⋅92⋅9−52−4⋅114⋅9−52−4⋅74⋅912−−19]=[00−6718−59181118]\left[\begin{matrix}- \frac{0 \cdot 4}{9} & 2 - \frac{4 \cdot 9}{2 \cdot 9} & - \frac{5}{2} - \frac{4 \cdot 11}{4 \cdot 9} & - \frac{5}{2} - \frac{4 \cdot 7}{4 \cdot 9} & \frac{1}{2} - - \frac{1}{9}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\end{matrix}\right]
obtenemos
[40−29−34910909211474−1400−6718−591811180−72514474−94]\left[\begin{matrix}4 & 0 & - \frac{2}{9} & - \frac{34}{9} & \frac{10}{9}\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\\0 & - \frac{7}{2} & \frac{51}{4} & \frac{47}{4} & - \frac{9}{4}\end{matrix}\right]
De 4 de fila restamos:
[−(−7)09−72−(−7)92⋅9514−(−7)114⋅9474−(−7)74⋅9−94−−−736]=[0013491189−229]\left[\begin{matrix}- \frac{\left(-7\right) 0}{9} & - \frac{7}{2} - \frac{\left(-7\right) 9}{2 \cdot 9} & \frac{51}{4} - \frac{\left(-7\right) 11}{4 \cdot 9} & \frac{47}{4} - \frac{\left(-7\right) 7}{4 \cdot 9} & - \frac{9}{4} - - \frac{-7}{36}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{134}{9} & \frac{118}{9} & - \frac{22}{9}\end{matrix}\right]
obtenemos
[40−29−34910909211474−1400−6718−591811180013491189−229]\left[\begin{matrix}4 & 0 & - \frac{2}{9} & - \frac{34}{9} & \frac{10}{9}\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\\0 & 0 & \frac{134}{9} & \frac{118}{9} & - \frac{22}{9}\end{matrix}\right]
En 3 de columna
[−29114−67181349]\left[\begin{matrix}- \frac{2}{9}\\\frac{11}{4}\\- \frac{67}{18}\\\frac{134}{9}\end{matrix}\right]
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
[00−6718−59181118]\left[\begin{matrix}0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[4−0⋅467−0⋅467−29−−29−349−−118603109−4⋅1118⋅67]=[400−240677267]\left[\begin{matrix}4 - \frac{0 \cdot 4}{67} & - \frac{0 \cdot 4}{67} & - \frac{2}{9} - - \frac{2}{9} & - \frac{34}{9} - - \frac{118}{603} & \frac{10}{9} - \frac{4 \cdot 11}{18 \cdot 67}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 0 & - \frac{240}{67} & \frac{72}{67}\end{matrix}\right]
obtenemos
[400−24067726709211474−1400−6718−591811180013491189−229]\left[\begin{matrix}4 & 0 & 0 & - \frac{240}{67} & \frac{72}{67}\\0 & \frac{9}{2} & \frac{11}{4} & \frac{7}{4} & - \frac{1}{4}\\0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\\0 & 0 & \frac{134}{9} & \frac{118}{9} & - \frac{22}{9}\end{matrix}\right]
De 2 de fila restamos:
[−(−99)013492−(−99)0134114−−−11474−−−649268−14−(−99)1118⋅134]=[0920−456727134]\left[\begin{matrix}- \frac{\left(-99\right) 0}{134} & \frac{9}{2} - \frac{\left(-99\right) 0}{134} & \frac{11}{4} - - \frac{-11}{4} & \frac{7}{4} - - \frac{-649}{268} & - \frac{1}{4} - \frac{\left(-99\right) 11}{18 \cdot 134}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{2} & 0 & - \frac{45}{67} & \frac{27}{134}\end{matrix}\right]
obtenemos
[400−2406772670920−45672713400−6718−591811180013491189−229]\left[\begin{matrix}4 & 0 & 0 & - \frac{240}{67} & \frac{72}{67}\\0 & \frac{9}{2} & 0 & - \frac{45}{67} & \frac{27}{134}\\0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\\0 & 0 & \frac{134}{9} & \frac{118}{9} & - \frac{22}{9}\end{matrix}\right]
De 4 de fila restamos:
[−(−4)0−(−4)01349−−−13491189−−−1189−229−−229]=[00000]\left[\begin{matrix}- \left(-4\right) 0 & - \left(-4\right) 0 & \frac{134}{9} - - \frac{-134}{9} & \frac{118}{9} - - \frac{-118}{9} & - \frac{22}{9} - - \frac{22}{9}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0 & 0 & 0\end{matrix}\right]
obtenemos
[400−2406772670920−45672713400−6718−5918111800000]\left[\begin{matrix}4 & 0 & 0 & - \frac{240}{67} & \frac{72}{67}\\0 & \frac{9}{2} & 0 & - \frac{45}{67} & \frac{27}{134}\\0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\\0 & 0 & 0 & 0 & 0\end{matrix}\right]
En 1 de columna
[4000]\left[\begin{matrix}4\\0\\0\\0\end{matrix}\right]
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
[400−240677267]\left[\begin{matrix}4 & 0 & 0 & - \frac{240}{67} & \frac{72}{67}\end{matrix}\right]
,
y lo restaremos de otras filas:
En 2 de columna
[09200]\left[\begin{matrix}0\\\frac{9}{2}\\0\\0\end{matrix}\right]
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
[0920−456727134]\left[\begin{matrix}0 & \frac{9}{2} & 0 & - \frac{45}{67} & \frac{27}{134}\end{matrix}\right]
,
y lo restaremos de otras filas:
En 3 de columna
[00−67180]\left[\begin{matrix}0\\0\\- \frac{67}{18}\\0\end{matrix}\right]
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
[00−6718−59181118]\left[\begin{matrix}0 & 0 & - \frac{67}{18} & - \frac{59}{18} & \frac{11}{18}\end{matrix}\right]
,
y lo restaremos de otras filas:

Todo está casi listo, sólo hace falta encontrar la incógnita, resolviendo las ecuaciones ordinarias:
4x1−240x467−7267=04 x_{1} - \frac{240 x_{4}}{67} - \frac{72}{67} = 0
9x22−45x467−27134=0\frac{9 x_{2}}{2} - \frac{45 x_{4}}{67} - \frac{27}{134} = 0
−67x318−59x418−1118=0- \frac{67 x_{3}}{18} - \frac{59 x_{4}}{18} - \frac{11}{18} = 0
0−0=00 - 0 = 0
Obtenemos como resultado:
x1=60x467+1867x_{1} = \frac{60 x_{4}}{67} + \frac{18}{67}
x2=10x467+367x_{2} = \frac{10 x_{4}}{67} + \frac{3}{67}
x3=−59x467−1167x_{3} = - \frac{59 x_{4}}{67} - \frac{11}{67}
donde x4 - variables libres
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