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Solución de sistemas de ecuaciones/ 2x1+3x2+4x3=0; 6x1-2x2+x3=0; -3x1+3x2+x3=0

2x1+3x2+4x3=0; 6x1-2x2+x3=0; -3x1+3x2+x3=0

v

Gráfico:

interior superior

interior superior

Solución

Ha introducido [src] 
2*x1 + 3*x2 + 4*x3 = 0
4x3+(2x1+3x2)=04 x_{3} + \left(2 x_{1} + 3 x_{2}\right) = 0 
6*x1 - 2*x2 + x3 = 0
x3+(6x1−2x2)=0x_{3} + \left(6 x_{1} - 2 x_{2}\right) = 0 
-3*x1 + 3*x2 + x3 = 0
x3+(−3x1+3x2)=0x_{3} + \left(- 3 x_{1} + 3 x_{2}\right) = 0 
x3 - 3*x1 + 3*x2 = 0
Respuesta rápida
x31=0x_{31} = 0
=
00
=
0

x21=0x_{21} = 0
=
00
=
0

x11=0x_{11} = 0
=
00
=
0
Método de Gauss
Tenemos el sistema de ecuaciones
4x3+(2x1+3x2)=04 x_{3} + \left(2 x_{1} + 3 x_{2}\right) = 0
x3+(6x1−2x2)=0x_{3} + \left(6 x_{1} - 2 x_{2}\right) = 0
x3+(−3x1+3x2)=0x_{3} + \left(- 3 x_{1} + 3 x_{2}\right) = 0

Expresamos el sistema de ecuaciones en su forma canónica
2x1+3x2+4x3=02 x_{1} + 3 x_{2} + 4 x_{3} = 0
6x1−2x2+x3=06 x_{1} - 2 x_{2} + x_{3} = 0
−3x1+3x2+x3=0- 3 x_{1} + 3 x_{2} + x_{3} = 0
Presentamos el sistema de ecuaciones lineales como matriz
[23406−210−3310]\left[\begin{matrix}2 & 3 & 4 & 0\\6 & -2 & 1 & 0\\-3 & 3 & 1 & 0\end{matrix}\right]
En 1 de columna
[26−3]\left[\begin{matrix}2\\6\\-3\end{matrix}\right]
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
[2340]\left[\begin{matrix}2 & 3 & 4 & 0\end{matrix}\right]
,
y lo restaremos de otras filas:
De 2 de fila restamos:
[6−2⋅3−3⋅3−21−3⋅4−0⋅3]=[0−11−110]\left[\begin{matrix}6 - 2 \cdot 3 & - 3 \cdot 3 - 2 & 1 - 3 \cdot 4 & - 0 \cdot 3\end{matrix}\right] = \left[\begin{matrix}0 & -11 & -11 & 0\end{matrix}\right]
obtenemos
[23400−11−110−3310]\left[\begin{matrix}2 & 3 & 4 & 0\\0 & -11 & -11 & 0\\-3 & 3 & 1 & 0\end{matrix}\right]
De 3 de fila restamos:
[−3−(−3)223−(−3)321−(−3)42−(−3)02]=[015270]\left[\begin{matrix}-3 - \frac{\left(-3\right) 2}{2} & 3 - \frac{\left(-3\right) 3}{2} & 1 - \frac{\left(-3\right) 4}{2} & - \frac{\left(-3\right) 0}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{15}{2} & 7 & 0\end{matrix}\right]
obtenemos
[23400−11−110015270]\left[\begin{matrix}2 & 3 & 4 & 0\\0 & -11 & -11 & 0\\0 & \frac{15}{2} & 7 & 0\end{matrix}\right]
En 2 de columna
[3−11152]\left[\begin{matrix}3\\-11\\\frac{15}{2}\end{matrix}\right]
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
[0−11−110]\left[\begin{matrix}0 & -11 & -11 & 0\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[2−(−3)0113−−−34−−−3−(−3)011]=[2010]\left[\begin{matrix}2 - \frac{\left(-3\right) 0}{11} & 3 - - -3 & 4 - - -3 & - \frac{\left(-3\right) 0}{11}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 1 & 0\end{matrix}\right]
obtenemos
[20100−11−110015270]\left[\begin{matrix}2 & 0 & 1 & 0\\0 & -11 & -11 & 0\\0 & \frac{15}{2} & 7 & 0\end{matrix}\right]
De 3 de fila restamos:
[−(−15)022152−−−1527−−−152−(−15)022]=[00−120]\left[\begin{matrix}- \frac{\left(-15\right) 0}{22} & \frac{15}{2} - - \frac{-15}{2} & 7 - - \frac{-15}{2} & - \frac{\left(-15\right) 0}{22}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{1}{2} & 0\end{matrix}\right]
obtenemos
[20100−11−11000−120]\left[\begin{matrix}2 & 0 & 1 & 0\\0 & -11 & -11 & 0\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right]
En 3 de columna
[1−11−12]\left[\begin{matrix}1\\-11\\- \frac{1}{2}\end{matrix}\right]
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
[00−120]\left[\begin{matrix}0 & 0 & - \frac{1}{2} & 0\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[2−(−2)0−(−2)01−−−1−(−2)0]=[2000]\left[\begin{matrix}2 - \left(-2\right) 0 & - \left(-2\right) 0 & 1 - - -1 & - \left(-2\right) 0\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 0\end{matrix}\right]
obtenemos
[20000−11−11000−120]\left[\begin{matrix}2 & 0 & 0 & 0\\0 & -11 & -11 & 0\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right]
De 2 de fila restamos:
[−0⋅22−11−0⋅22−11−(−1)222−0⋅22]=[0−1100]\left[\begin{matrix}- 0 \cdot 22 & -11 - 0 \cdot 22 & -11 - \frac{\left(-1\right) 22}{2} & - 0 \cdot 22\end{matrix}\right] = \left[\begin{matrix}0 & -11 & 0 & 0\end{matrix}\right]
obtenemos
[20000−110000−120]\left[\begin{matrix}2 & 0 & 0 & 0\\0 & -11 & 0 & 0\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right]

Todo está casi listo, sólo hace falta encontrar la incógnita, resolviendo las ecuaciones ordinarias:
2x1=02 x_{1} = 0
−11x2=0- 11 x_{2} = 0
−x32=0- \frac{x_{3}}{2} = 0
Obtenemos como resultado:
x1=0x_{1} = 0
x2=0x_{2} = 0
x3=0x_{3} = 0
Regla de Cramer
4x3+(2x1+3x2)=04 x_{3} + \left(2 x_{1} + 3 x_{2}\right) = 0
x3+(6x1−2x2)=0x_{3} + \left(6 x_{1} - 2 x_{2}\right) = 0
x3+(−3x1+3x2)=0x_{3} + \left(- 3 x_{1} + 3 x_{2}\right) = 0

Expresamos el sistema de ecuaciones en su forma canónica
2x1+3x2+4x3=02 x_{1} + 3 x_{2} + 4 x_{3} = 0
6x1−2x2+x3=06 x_{1} - 2 x_{2} + x_{3} = 0
−3x1+3x2+x3=0- 3 x_{1} + 3 x_{2} + x_{3} = 0
Presentamos el sistema de ecuaciones lineales como matriz
[2x1+3x2+4x36x1−2x2+x3−3x1+3x2+x3]=[000]\left[\begin{matrix}2 x_{1} + 3 x_{2} + 4 x_{3}\\6 x_{1} - 2 x_{2} + x_{3}\\- 3 x_{1} + 3 x_{2} + x_{3}\end{matrix}\right] = \left[\begin{matrix}0\\0\\0\end{matrix}\right]
- es el sistema de ecuaciones en forma de
A*x = B

De la siguiente forma resolvemos una ecuación matriz de este tipo aplicando la regla de Cramer:

Como el determinante de la matriz:
A=det⁡([2346−21−331])=11A = \operatorname{det}{\left(\left[\begin{matrix}2 & 3 & 4\\6 & -2 & 1\\-3 & 3 & 1\end{matrix}\right] \right)} = 11
, entonces
Raíz xi obtenemos dividiendo el determinador de la matriz Ai. por el determinador de la matriz A.
( Ai obtenemos sustituyendo en la matriz A de columna i por columna B )
x1=det⁡([0340−21031])11=0x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}0 & 3 & 4\\0 & -2 & 1\\0 & 3 & 1\end{matrix}\right] \right)}}{11} = 0
x2=det⁡([204601−301])11=0x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 0 & 4\\6 & 0 & 1\\-3 & 0 & 1\end{matrix}\right] \right)}}{11} = 0
x3=det⁡([2306−20−330])11=0x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 3 & 0\\6 & -2 & 0\\-3 & 3 & 0\end{matrix}\right] \right)}}{11} = 0
Respuesta numérica [src] 
x11 = -1.550963648536927e-25
x21 = -2.584939414228211e-25
x31 = 3.360421238496675e-25
x12 = 0
x22 = 2.067951531382569e-25
x32 = -2.067951531382569e-25
x13 = -1.033975765691285e-25
x23 = -2.067951531382569e-25
x33 = 2.067951531382569e-25
x14 = -2.067951531382569e-25
x24 = -4.135903062765138e-25
x34 = 5.169878828456423e-25
x15 = 2.067951531382569e-25
x25 = 4.135903062765138e-25
x35 = -5.686866711302065e-25
x16 = -3.101927297073854e-25
x26 = -6.203854594147708e-25
x36 = 8.271806125530277e-25
x17 = 1.550963648536927e-25
x27 = 2.584939414228211e-25
x37 = -3.360421238496675e-25
x18 = 0
x28 = 1.033975765691285e-25
x38 = -1.033975765691285e-25
x19 = -3.360421238496675e-25
x29 = -6.203854594147708e-25
x39 = 8.271806125530277e-25
x110 = 4.135903062765138e-25
x210 = 8.271806125530277e-25
x310 = -1.033975765691285e-24
x111 = -7.754818242684634e-26
x211 = -1.033975765691285e-25
x311 = 1.033975765691285e-25
x112 = -1.809457589959748e-25
x212 = -3.101927297073854e-25
x312 = 4.135903062765138e-25
x113 = -1.033975765691285e-24
x213 = -2.067951531382569e-24
x313 = 2.68833699079734e-24
x114 = -4.135903062765138e-25
x214 = -6.203854594147708e-25
x314 = 8.271806125530277e-25
x115 = 0
x215 = -1.033975765691285e-25
x315 = 1.033975765691285e-25
x116 = -2.067951531382569e-25
x216 = -3.101927297073854e-25
x316 = 4.135903062765138e-25
x117 = 0
x217 = 2.067951531382569e-25
x317 = -3.360421238496675e-25
x118 = 0
x218 = -2.067951531382569e-25
x318 = 2.067951531382569e-25
x119 = 0
x219 = 7.754818242684634e-26
x319 = -1.033975765691285e-25
x120 = 0
x220 = 2.067951531382569e-25
x320 = -3.101927297073854e-25
x121 = 2.067951531382569e-25
x221 = 2.584939414228211e-25
x321 = -3.360421238496675e-25
x122 = 1.033975765691285e-25
x222 = 2.584939414228211e-25
x322 = -3.360421238496675e-25
x123 = -3.101927297073854e-25
x223 = -3.101927297073854e-25
x323 = 4.135903062765138e-25
x124 = 1.033975765691285e-25
x224 = 2.067951531382569e-25
x324 = -2.067951531382569e-25
x125 = -3.618915179919496e-25
x225 = -6.203854594147708e-25
x325 = 8.271806125530277e-25
x126 = 0
x226 = -2.584939414228211e-26
x326 = 2.584939414228211e-26
x127 = 1.033975765691285e-24
x227 = 2.067951531382569e-24
x327 = -2.68833699079734e-24
x127 = 1.033975765691285e-24
x227 = 2.067951531382569e-24
x327 = -2.68833699079734e-24
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