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Solución de sistemas de ecuaciones/ pi*x/2+y*pi^2/8+z*pi^3/24=1; x*pi^2/8+y*pi^3/24+z*pi^4/64=pi/2-1; x*pi^3/24+y*pi^4/64+z*pi^5/160=pi^2/4-1

pi*x/2+y*pi^2/8+z*pi^3/24=1; x*pi^2/8+y*pi^3/24+z*pi^4/64=pi/2-1; x*pi^3/24+y*pi^4/64+z*pi^5/160=pi^2/4-1

v

Gráfico:

interior superior

interior superior

Solución

Ha introducido [src] 
           2       3    
pi*x   y*pi    z*pi     
---- + ----- + ----- = 1
 2       8       24
π3z24+(πx2+π2y8)=1\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1 
    2       3       4         
x*pi    y*pi    z*pi    pi    
----- + ----- + ----- = -- - 1
  8       24      64    2
π4z64+(π2x8+π3y24)=−1+π2\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2} 
    3       4       5     2    
x*pi    y*pi    z*pi    pi     
----- + ----- + ----- = --- - 1
  24      64     160     4
π5z160+(π3x24+π4y64)=−1+π24\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4} 
(pi^5*z)/160 + (pi^3*x)/24 + (pi^4*y)/64 = -1 + pi^2/4
Respuesta rápida
x1=6(−40+π2+24π)π3x_{1} = \frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}
=
6(−40+π2+24π)π3\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}
=
8.75974149763151

y1=−1536π−96π2+2880π4y_{1} = \frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}
=
96(−16π−π2+30)π4\frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}
=
-29.699161625568

z1=240(−24+π2+12π)π5z_{1} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}
=
240(−24+π2+12π)π5\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}
=
18.4840785734713
Regla de Cramer
π3z24+(πx2+π2y8)=1\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1
π4z64+(π2x8+π3y24)=−1+π2\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}
π5z160+(π3x24+π4y64)=−1+π24\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}

Expresamos el sistema de ecuaciones en su forma canónica
πx2+π2y8+π3z24−1=0\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0
π2x8+π3y24+π4z64−π2+1=0\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0
π3x24+π4y64+π5z160−π24+1=0\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0
Presentamos el sistema de ecuaciones lineales como matriz
[π2x1+π28x2+π324x3π28x1+π324x2+π464x3π324x1+π464x2+π5160x3]=[1−1+π2−1+π24]\left[\begin{matrix}\frac{\pi}{2} x_{1} + \frac{\pi^{2}}{8} x_{2} + \frac{\pi^{3}}{24} x_{3}\\\frac{\pi^{2}}{8} x_{1} + \frac{\pi^{3}}{24} x_{2} + \frac{\pi^{4}}{64} x_{3}\\\frac{\pi^{3}}{24} x_{1} + \frac{\pi^{4}}{64} x_{2} + \frac{\pi^{5}}{160} x_{3}\end{matrix}\right] = \left[\begin{matrix}1\\-1 + \frac{\pi}{2}\\-1 + \frac{\pi^{2}}{4}\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⁡([π2π28π324π28π324π464π324π464π5160])=π91105920A = \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)} = \frac{\pi^{9}}{1105920}
, 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=1105920det⁡([1π28π324−1+π2π324π464−1+π24π464π5160])π9=2(−240(−1−π(−1+π4)2+π26)π2+1−12(−30(−1−π(−1+π4)2+π26)π−1+π4)π)πx_{1} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}1 & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\-1 + \frac{\pi}{2} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\-1 + \frac{\pi^{2}}{4} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{2 \left(- \frac{240 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} + 1 - \frac{12 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi}\right)}{\pi}
=
6(−40+π2+24π)π3\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}
x2=1105920det⁡([π21π324π28−1+π2π464π324−1+π24π5160])π9=96(−30(−1−π(−1+π4)2+π26)π−1+π4)π3x_{2} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & 1 & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & -1 + \frac{\pi}{2} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & -1 + \frac{\pi^{2}}{4} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{96 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi^{3}}
=
−1536π−96π2+2880π4\frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}
x3=1105920det⁡([π2π281π28π324−1+π2π324π464−1+π24])π9=5760(−1−π(−1+π4)2+π26)π5x_{3} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{5760 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{5}}
=
240(−24+π2+12π)π5\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}
Método de Gauss
Tenemos el sistema de ecuaciones
π3z24+(πx2+π2y8)=1\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1
π4z64+(π2x8+π3y24)=−1+π2\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}
π5z160+(π3x24+π4y64)=−1+π24\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}

Expresamos el sistema de ecuaciones en su forma canónica
πx2+π2y8+π3z24−1=0\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0
π2x8+π3y24+π4z64−π2+1=0\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0
π3x24+π4y64+π5z160−π24+1=0\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0
Presentamos el sistema de ecuaciones lineales como matriz
[π2π28π3241π28π324π464−1+π2π324π464π5160−1+π24]\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]
En 1 de columna
[π2π28π324]\left[\begin{matrix}\frac{\pi}{2}\\\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{24}\end{matrix}\right]
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
[π2π28π3241]\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\end{matrix}\right]
,
y lo restaremos de otras filas:
De 2 de fila restamos:
[−π4π2+π28−π4π28+π324−π4π324+π464−π4+(−1+π2)]=[0π396π4192−1+π4]\left[\begin{matrix}- \frac{\pi}{4} \frac{\pi}{2} + \frac{\pi^{2}}{8} & - \frac{\pi}{4} \frac{\pi^{2}}{8} + \frac{\pi^{3}}{24} & - \frac{\pi}{4} \frac{\pi^{3}}{24} + \frac{\pi^{4}}{64} & - \frac{\pi}{4} + \left(-1 + \frac{\pi}{2}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]
obtenemos
[π2π28π32410π396π4192−1+π4π324π464π5160−1+π24]\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]
De 3 de fila restamos:
[−π2π212+π324−π212π28+π464−π212π324+π5160−π212+(−1+π24)]=[0π4192π5360−1+π26]\left[\begin{matrix}- \frac{\pi}{2} \frac{\pi^{2}}{12} + \frac{\pi^{3}}{24} & - \frac{\pi^{2}}{12} \frac{\pi^{2}}{8} + \frac{\pi^{4}}{64} & - \frac{\pi^{2}}{12} \frac{\pi^{3}}{24} + \frac{\pi^{5}}{160} & - \frac{\pi^{2}}{12} + \left(-1 + \frac{\pi^{2}}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]
obtenemos
[π2π28π32410π396π4192−1+π40π4192π5360−1+π26]\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]
En 2 de columna
[π28π396π4192]\left[\begin{matrix}\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{96}\\\frac{\pi^{4}}{192}\end{matrix}\right]
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
[0π396π4192−1+π4]\left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[−012π+π2−12ππ396+π28−12ππ4192+π3241−12π(−1+π4)]=[π20−π348−12(−1+π4)π+1]\left[\begin{matrix}- 0 \frac{12}{\pi} + \frac{\pi}{2} & - \frac{12}{\pi} \frac{\pi^{3}}{96} + \frac{\pi^{2}}{8} & - \frac{12}{\pi} \frac{\pi^{4}}{192} + \frac{\pi^{3}}{24} & 1 - \frac{12}{\pi} \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\end{matrix}\right]
obtenemos
[π20−π348−12(−1+π4)π+10π396π4192−1+π40π4192π5360−1+π26]\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]
De 3 de fila restamos:
[−0π2−π2π396+π4192−π2π4192+π5360−π2(−1+π4)+(−1+π26)]=[00π55760−1−π(−1+π4)2+π26]\left[\begin{matrix}- 0 \frac{\pi}{2} & - \frac{\pi}{2} \frac{\pi^{3}}{96} + \frac{\pi^{4}}{192} & - \frac{\pi}{2} \frac{\pi^{4}}{192} + \frac{\pi^{5}}{360} & - \frac{\pi}{2} \left(-1 + \frac{\pi}{4}\right) + \left(-1 + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]
obtenemos
[π20−π348−12(−1+π4)π+10π396π4192−1+π400π55760−1−π(−1+π4)2+π26]\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]
En 3 de columna
[−π348π4192π55760]\left[\begin{matrix}- \frac{\pi^{3}}{48}\\\frac{\pi^{4}}{192}\\\frac{\pi^{5}}{5760}\end{matrix}\right]
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
[00π55760−1−π(−1+π4)2+π26]\left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]
,
y lo restaremos de otras filas:
De 1 de fila restamos:
[−0(−120π2)+π2−0(−120π2)−π348−−120π2π55760(−12(−1+π4)π+1)−−120π2(−1−π(−1+π4)2+π26)]=[π200−12(−1+π4)π+1+120(−1−π(−1+π4)2+π26)π2]\left[\begin{matrix}- 0 \left(- \frac{120}{\pi^{2}}\right) + \frac{\pi}{2} & - 0 \left(- \frac{120}{\pi^{2}}\right) & - \frac{\pi^{3}}{48} - - \frac{120}{\pi^{2}} \frac{\pi^{5}}{5760} & \left(- \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\right) - - \frac{120}{\pi^{2}} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\end{matrix}\right]
obtenemos
[π200−12(−1+π4)π+1+120(−1−π(−1+π4)2+π26)π20π396π4192−1+π400π55760−1−π(−1+π4)2+π26]\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]
De 2 de fila restamos:
[−030π−030π+π396−30ππ55760+π4192−30π(−1−π(−1+π4)2+π26)+(−1+π4)]=[0π3960−30(−1−π(−1+π4)2+π26)π−1+π4]\left[\begin{matrix}- 0 \frac{30}{\pi} & - 0 \frac{30}{\pi} + \frac{\pi^{3}}{96} & - \frac{30}{\pi} \frac{\pi^{5}}{5760} + \frac{\pi^{4}}{192} & - \frac{30}{\pi} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right) + \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\end{matrix}\right]
obtenemos
[π200−12(−1+π4)π+1+120(−1−π(−1+π4)2+π26)π20π3960−30(−1−π(−1+π4)2+π26)π−1+π400π55760−1−π(−1+π4)2+π26]\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]

Todo está casi listo, sólo hace falta encontrar la incógnita, resolviendo las ecuaciones ordinarias:
πx12−120(−1−π(−1+π4)2+π26)π2−1+12(−1+π4)π=0\frac{\pi x_{1}}{2} - \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} - 1 + \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} = 0
π3x296−π4+1+30(−1−π(−1+π4)2+π26)π=0\frac{\pi^{3} x_{2}}{96} - \frac{\pi}{4} + 1 + \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} = 0
π5x35760−π26+π(−1+π4)2+1=0\frac{\pi^{5} x_{3}}{5760} - \frac{\pi^{2}}{6} + \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + 1 = 0
Obtenemos como resultado:
x1=6(−40+π2+24π)π3x_{1} = \frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}
x2=96(−16π−π2+30)π4x_{2} = \frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}
x3=240(−24+π2+12π)π5x_{3} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}
Respuesta numérica [src] 
x1 = 8.759741497631506
y1 = -29.69916162556796
z1 = 18.48407857347126
x1 = 8.759741497631506
y1 = -29.69916162556796
z1 = 18.48407857347126
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