Respuesta:
La solución del sistema es x = 120/37, y = -29/37, z = 10/37
Explicación paso a paso:
Método por determinantes (Regla de Sarrus):
4x+4y+8z=12
4x-5y-7z=15
3x+2y-8z=6
Ahora calculamos el determinante auxiliar:
[tex]|A|= \left[\begin{array}{ccc}4&4&8\\4&-5&-7\\3&2&-8\end{array}\right] = (4)(-5)(-8)+(4)(2)(8)+(4)(-7)(3)-(3)(-5)(8)-(2)(-7)(4)-(4)(4)(-8) =160+64-84+120+56+128=444[/tex]
Ahora calculamos el determinante auxiliar en x:
[tex]|A_x|= \left[\begin{array}{ccc}12&4&8\\15&-5&-7\\6&2&-8\end{array}\right] = (12)(-5)(-8)+(15)(2)(8)+(4)(-7)(6)-(6)(-5)(8)-(2)(-7)(12)-(15)(4)(-8) =480+240-168+240+168+480=1440[/tex]
Ahora calculamos el determinante auxiliar en y:
[tex]|A_y|= \left[\begin{array}{ccc}4&12&8\\4&15&-7\\3&6&-8\end{array}\right] = (4)(15)(-8)+(4)(6)(8)+(12)(-7)(3)-(3)(15)(8)-(6)(-7)(4)-(4)(12)(-8) =-480+192-252-360+168+384=-348[/tex]
Y finalmente calculamos el determinante auxiliar en z:
[tex]|A_z|= \left[\begin{array}{ccc}4&4&12\\4&-5&15\\3&2&6\end{array}\right] = (4)(-5)(6)+(4)(2)(12)+(4)(15)(3)-(3)(-5)(12)-(2)(15)(4)-(4)(4)(6) =-120+96+180+180-120-96=120[/tex]
Ahora podemos calcular la solución:
[tex]x = \frac{|A_x|}{A} = \frac{1440}{444} =\frac{120}{37}[/tex]
[tex]y = \frac{|A_y|}{A} = \frac{-348}{444} =\frac{-29}{37}[/tex]
[tex]z = \frac{|A_z|}{A} = \frac{120}{444} =\frac{10}{37}[/tex]
Por lo tanto, la solución del sistema es x = 120/37, y = -29/37, z = 10/37
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
Respuesta:
La solución del sistema es x = 120/37, y = -29/37, z = 10/37
Explicación paso a paso:
Método por determinantes (Regla de Sarrus):
4x+4y+8z=12
4x-5y-7z=15
3x+2y-8z=6
Ahora calculamos el determinante auxiliar:
[tex]|A|= \left[\begin{array}{ccc}4&4&8\\4&-5&-7\\3&2&-8\end{array}\right] = (4)(-5)(-8)+(4)(2)(8)+(4)(-7)(3)-(3)(-5)(8)-(2)(-7)(4)-(4)(4)(-8) =160+64-84+120+56+128=444[/tex]
Ahora calculamos el determinante auxiliar en x:
[tex]|A_x|= \left[\begin{array}{ccc}12&4&8\\15&-5&-7\\6&2&-8\end{array}\right] = (12)(-5)(-8)+(15)(2)(8)+(4)(-7)(6)-(6)(-5)(8)-(2)(-7)(12)-(15)(4)(-8) =480+240-168+240+168+480=1440[/tex]
Ahora calculamos el determinante auxiliar en y:
[tex]|A_y|= \left[\begin{array}{ccc}4&12&8\\4&15&-7\\3&6&-8\end{array}\right] = (4)(15)(-8)+(4)(6)(8)+(12)(-7)(3)-(3)(15)(8)-(6)(-7)(4)-(4)(12)(-8) =-480+192-252-360+168+384=-348[/tex]
Y finalmente calculamos el determinante auxiliar en z:
[tex]|A_z|= \left[\begin{array}{ccc}4&4&12\\4&-5&15\\3&2&6\end{array}\right] = (4)(-5)(6)+(4)(2)(12)+(4)(15)(3)-(3)(-5)(12)-(2)(15)(4)-(4)(4)(6) =-120+96+180+180-120-96=120[/tex]
Ahora podemos calcular la solución:
[tex]x = \frac{|A_x|}{A} = \frac{1440}{444} =\frac{120}{37}[/tex]
[tex]y = \frac{|A_y|}{A} = \frac{-348}{444} =\frac{-29}{37}[/tex]
[tex]z = \frac{|A_z|}{A} = \frac{120}{444} =\frac{10}{37}[/tex]
Por lo tanto, la solución del sistema es x = 120/37, y = -29/37, z = 10/37