Sr Examen
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
Solución
Ha introducido
[src]
2 3 pi*x y*pi z*pi ---- + ----- + ----- = 1 2 8 24
$$\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1$$
2 3 4 x*pi y*pi z*pi pi ----- + ----- + ----- = -- - 1 8 24 64 2
$$\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}$$
3 4 5 2 x*pi y*pi z*pi pi ----- + ----- + ----- = --- - 1 24 64 160 4
$$\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}$$
(pi^5*z)/160 + (pi^3*x)/24 + (pi^4*y)/64 = -1 + pi^2/4
Respuesta rápida
$$x_{1} = \frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
=
$$\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
=
$$y_{1} = \frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}$$
=
$$\frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}$$
=
$$z_{1} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
=
$$\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
=
=
$$\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
=
8.75974149763151
$$y_{1} = \frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}$$
=
$$\frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}$$
=
-29.699161625568
$$z_{1} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
=
$$\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
=
18.4840785734713
Regla de Cramer
$$\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1$$
$$\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}$$
$$\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}$$
Expresamos el sistema de ecuaciones en su forma canónica
$$\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0$$
$$\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0$$
$$\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0$$
Presentamos el sistema de ecuaciones lineales como matriz
$$\left[\begin{matrix}\frac{\pi}{2} x_{1} + \frac{\pi^{2}}{8} x_{2} + \frac{\pi^{3}}{24} x_{3}\\\frac{\pi^{2}}{8} x_{1} + \frac{\pi^{3}}{24} x_{2} + \frac{\pi^{4}}{64} x_{3}\\\frac{\pi^{3}}{24} x_{1} + \frac{\pi^{4}}{64} x_{2} + \frac{\pi^{5}}{160} x_{3}\end{matrix}\right] = \left[\begin{matrix}1\\-1 + \frac{\pi}{2}\\-1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)} = \frac{\pi^{9}}{1105920}$$
, 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 )
$$x_{1} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}1 & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\-1 + \frac{\pi}{2} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\-1 + \frac{\pi^{2}}{4} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{2 \left(- \frac{240 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} + 1 - \frac{12 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi}\right)}{\pi}$$
=
$$\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
$$x_{2} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & 1 & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & -1 + \frac{\pi}{2} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & -1 + \frac{\pi^{2}}{4} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{96 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi^{3}}$$
=
$$\frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}$$
$$x_{3} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{5760 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{5}}$$
=
$$\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
$$\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}$$
$$\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}$$
Expresamos el sistema de ecuaciones en su forma canónica
$$\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0$$
$$\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0$$
$$\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0$$
Presentamos el sistema de ecuaciones lineales como matriz
$$\left[\begin{matrix}\frac{\pi}{2} x_{1} + \frac{\pi^{2}}{8} x_{2} + \frac{\pi^{3}}{24} x_{3}\\\frac{\pi^{2}}{8} x_{1} + \frac{\pi^{3}}{24} x_{2} + \frac{\pi^{4}}{64} x_{3}\\\frac{\pi^{3}}{24} x_{1} + \frac{\pi^{4}}{64} x_{2} + \frac{\pi^{5}}{160} x_{3}\end{matrix}\right] = \left[\begin{matrix}1\\-1 + \frac{\pi}{2}\\-1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)} = \frac{\pi^{9}}{1105920}$$
, 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 )
$$x_{1} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}1 & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24}\\-1 + \frac{\pi}{2} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64}\\-1 + \frac{\pi^{2}}{4} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{2 \left(- \frac{240 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} + 1 - \frac{12 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi}\right)}{\pi}$$
=
$$\frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
$$x_{2} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & 1 & \frac{\pi^{3}}{24}\\\frac{\pi^{2}}{8} & -1 + \frac{\pi}{2} & \frac{\pi^{4}}{64}\\\frac{\pi^{3}}{24} & -1 + \frac{\pi^{2}}{4} & \frac{\pi^{5}}{160}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{96 \left(- \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\right)}{\pi^{3}}$$
=
$$\frac{- 1536 \pi - 96 \pi^{2} + 2880}{\pi^{4}}$$
$$x_{3} = \frac{1105920 \operatorname{det}{\left(\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right] \right)}}{\pi^{9}} = \frac{5760 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{5}}$$
=
$$\frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
Método de Gauss
Tenemos el sistema de ecuaciones
$$\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1$$
$$\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}$$
$$\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}$$
Expresamos el sistema de ecuaciones en su forma canónica
$$\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0$$
$$\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0$$
$$\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0$$
Presentamos el sistema de ecuaciones lineales como matriz
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
En 1 de columna
$$\left[\begin{matrix}\frac{\pi}{2}\\\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{24}\end{matrix}\right]$$
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 2 de fila restamos:
$$\left[\begin{matrix}- \frac{\pi}{4} \frac{\pi}{2} + \frac{\pi^{2}}{8} & - \frac{\pi}{4} \frac{\pi^{2}}{8} + \frac{\pi^{3}}{24} & - \frac{\pi}{4} \frac{\pi^{3}}{24} + \frac{\pi^{4}}{64} & - \frac{\pi}{4} + \left(-1 + \frac{\pi}{2}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
De 3 de fila restamos:
$$\left[\begin{matrix}- \frac{\pi}{2} \frac{\pi^{2}}{12} + \frac{\pi^{3}}{24} & - \frac{\pi^{2}}{12} \frac{\pi^{2}}{8} + \frac{\pi^{4}}{64} & - \frac{\pi^{2}}{12} \frac{\pi^{3}}{24} + \frac{\pi^{5}}{160} & - \frac{\pi^{2}}{12} + \left(-1 + \frac{\pi^{2}}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
En 2 de columna
$$\left[\begin{matrix}\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{96}\\\frac{\pi^{4}}{192}\end{matrix}\right]$$
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
$$\left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 1 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{12}{\pi} + \frac{\pi}{2} & - \frac{12}{\pi} \frac{\pi^{3}}{96} + \frac{\pi^{2}}{8} & - \frac{12}{\pi} \frac{\pi^{4}}{192} + \frac{\pi^{3}}{24} & 1 - \frac{12}{\pi} \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
De 3 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{\pi}{2} & - \frac{\pi}{2} \frac{\pi^{3}}{96} + \frac{\pi^{4}}{192} & - \frac{\pi}{2} \frac{\pi^{4}}{192} + \frac{\pi^{5}}{360} & - \frac{\pi}{2} \left(-1 + \frac{\pi}{4}\right) + \left(-1 + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
En 3 de columna
$$\left[\begin{matrix}- \frac{\pi^{3}}{48}\\\frac{\pi^{4}}{192}\\\frac{\pi^{5}}{5760}\end{matrix}\right]$$
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
$$\left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 1 de fila restamos:
$$\left[\begin{matrix}- 0 \left(- \frac{120}{\pi^{2}}\right) + \frac{\pi}{2} & - 0 \left(- \frac{120}{\pi^{2}}\right) & - \frac{\pi^{3}}{48} - - \frac{120}{\pi^{2}} \frac{\pi^{5}}{5760} & \left(- \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\right) - - \frac{120}{\pi^{2}} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
De 2 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{30}{\pi} & - 0 \frac{30}{\pi} + \frac{\pi^{3}}{96} & - \frac{30}{\pi} \frac{\pi^{5}}{5760} + \frac{\pi^{4}}{192} & - \frac{30}{\pi} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right) + \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
Todo está casi listo, sólo hace falta encontrar la incógnita, resolviendo las ecuaciones ordinarias:
$$\frac{\pi x_{1}}{2} - \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} - 1 + \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} = 0$$
$$\frac{\pi^{3} x_{2}}{96} - \frac{\pi}{4} + 1 + \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} = 0$$
$$\frac{\pi^{5} x_{3}}{5760} - \frac{\pi^{2}}{6} + \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + 1 = 0$$
Obtenemos como resultado:
$$x_{1} = \frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
$$x_{2} = \frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}$$
$$x_{3} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
$$\frac{\pi^{3} z}{24} + \left(\frac{\pi x}{2} + \frac{\pi^{2} y}{8}\right) = 1$$
$$\frac{\pi^{4} z}{64} + \left(\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24}\right) = -1 + \frac{\pi}{2}$$
$$\frac{\pi^{5} z}{160} + \left(\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64}\right) = -1 + \frac{\pi^{2}}{4}$$
Expresamos el sistema de ecuaciones en su forma canónica
$$\frac{\pi x}{2} + \frac{\pi^{2} y}{8} + \frac{\pi^{3} z}{24} - 1 = 0$$
$$\frac{\pi^{2} x}{8} + \frac{\pi^{3} y}{24} + \frac{\pi^{4} z}{64} - \frac{\pi}{2} + 1 = 0$$
$$\frac{\pi^{3} x}{24} + \frac{\pi^{4} y}{64} + \frac{\pi^{5} z}{160} - \frac{\pi^{2}}{4} + 1 = 0$$
Presentamos el sistema de ecuaciones lineales como matriz
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\\frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & -1 + \frac{\pi}{2}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
En 1 de columna
$$\left[\begin{matrix}\frac{\pi}{2}\\\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{24}\end{matrix}\right]$$
hacemos que todos los elementos excepto
1 -del elemento son iguales a cero.
- Para ello, cogemos 1 fila
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 2 de fila restamos:
$$\left[\begin{matrix}- \frac{\pi}{4} \frac{\pi}{2} + \frac{\pi^{2}}{8} & - \frac{\pi}{4} \frac{\pi^{2}}{8} + \frac{\pi^{3}}{24} & - \frac{\pi}{4} \frac{\pi^{3}}{24} + \frac{\pi^{4}}{64} & - \frac{\pi}{4} + \left(-1 + \frac{\pi}{2}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\\frac{\pi^{3}}{24} & \frac{\pi^{4}}{64} & \frac{\pi^{5}}{160} & -1 + \frac{\pi^{2}}{4}\end{matrix}\right]$$
De 3 de fila restamos:
$$\left[\begin{matrix}- \frac{\pi}{2} \frac{\pi^{2}}{12} + \frac{\pi^{3}}{24} & - \frac{\pi^{2}}{12} \frac{\pi^{2}}{8} + \frac{\pi^{4}}{64} & - \frac{\pi^{2}}{12} \frac{\pi^{3}}{24} + \frac{\pi^{5}}{160} & - \frac{\pi^{2}}{12} + \left(-1 + \frac{\pi^{2}}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & \frac{\pi^{2}}{8} & \frac{\pi^{3}}{24} & 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
En 2 de columna
$$\left[\begin{matrix}\frac{\pi^{2}}{8}\\\frac{\pi^{3}}{96}\\\frac{\pi^{4}}{192}\end{matrix}\right]$$
hacemos que todos los elementos excepto
2 -del elemento son iguales a cero.
- Para ello, cogemos 2 fila
$$\left[\begin{matrix}0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 1 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{12}{\pi} + \frac{\pi}{2} & - \frac{12}{\pi} \frac{\pi^{3}}{96} + \frac{\pi^{2}}{8} & - \frac{12}{\pi} \frac{\pi^{4}}{192} + \frac{\pi^{3}}{24} & 1 - \frac{12}{\pi} \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & \frac{\pi^{4}}{192} & \frac{\pi^{5}}{360} & -1 + \frac{\pi^{2}}{6}\end{matrix}\right]$$
De 3 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{\pi}{2} & - \frac{\pi}{2} \frac{\pi^{3}}{96} + \frac{\pi^{4}}{192} & - \frac{\pi}{2} \frac{\pi^{4}}{192} + \frac{\pi^{5}}{360} & - \frac{\pi}{2} \left(-1 + \frac{\pi}{4}\right) + \left(-1 + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & - \frac{\pi^{3}}{48} & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
En 3 de columna
$$\left[\begin{matrix}- \frac{\pi^{3}}{48}\\\frac{\pi^{4}}{192}\\\frac{\pi^{5}}{5760}\end{matrix}\right]$$
hacemos que todos los elementos excepto
3 -del elemento son iguales a cero.
- Para ello, cogemos 3 fila
$$\left[\begin{matrix}0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
,
y lo restaremos de otras filas:
De 1 de fila restamos:
$$\left[\begin{matrix}- 0 \left(- \frac{120}{\pi^{2}}\right) + \frac{\pi}{2} & - 0 \left(- \frac{120}{\pi^{2}}\right) & - \frac{\pi^{3}}{48} - - \frac{120}{\pi^{2}} \frac{\pi^{5}}{5760} & \left(- \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1\right) - - \frac{120}{\pi^{2}} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)\end{matrix}\right] = \left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & \frac{\pi^{4}}{192} & -1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
De 2 de fila restamos:
$$\left[\begin{matrix}- 0 \frac{30}{\pi} & - 0 \frac{30}{\pi} + \frac{\pi^{3}}{96} & - \frac{30}{\pi} \frac{\pi^{5}}{5760} + \frac{\pi^{4}}{192} & - \frac{30}{\pi} \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right) + \left(-1 + \frac{\pi}{4}\right)\end{matrix}\right] = \left[\begin{matrix}0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\end{matrix}\right]$$
obtenemos
$$\left[\begin{matrix}\frac{\pi}{2} & 0 & 0 & - \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} + 1 + \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}}\\0 & \frac{\pi^{3}}{96} & 0 & - \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} - 1 + \frac{\pi}{4}\\0 & 0 & \frac{\pi^{5}}{5760} & -1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\end{matrix}\right]$$
Todo está casi listo, sólo hace falta encontrar la incógnita, resolviendo las ecuaciones ordinarias:
$$\frac{\pi x_{1}}{2} - \frac{120 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi^{2}} - 1 + \frac{12 \left(-1 + \frac{\pi}{4}\right)}{\pi} = 0$$
$$\frac{\pi^{3} x_{2}}{96} - \frac{\pi}{4} + 1 + \frac{30 \left(-1 - \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + \frac{\pi^{2}}{6}\right)}{\pi} = 0$$
$$\frac{\pi^{5} x_{3}}{5760} - \frac{\pi^{2}}{6} + \frac{\pi \left(-1 + \frac{\pi}{4}\right)}{2} + 1 = 0$$
Obtenemos como resultado:
$$x_{1} = \frac{6 \left(-40 + \pi^{2} + 24 \pi\right)}{\pi^{3}}$$
$$x_{2} = \frac{96 \left(- 16 \pi - \pi^{2} + 30\right)}{\pi^{4}}$$
$$x_{3} = \frac{240 \left(-24 + \pi^{2} + 12 \pi\right)}{\pi^{5}}$$
Respuesta numérica
[src]
x1 = 8.759741497631506 y1 = -29.69916162556796 z1 = 18.48407857347126
x1 = 8.759741497631506 y1 = -29.69916162556796 z1 = 18.48407857347126