Question

ayuda con este sistema de ecuaciones
x+y+mz=1
x+my+z=1
mx+y+z=-2

Answer 1

x+y+mz=1
x+my+z=1
mx+y+z=-2

x+y+mz=1
0x+(m-1)y+(1-m)z=0
0x+(1-m)y+(1-m²)z=-m-2

x+y+mz=1
0x+(m-1)y+(1-m)z=0
0y+0y+(2-m-m²)z=-m-2


x+y+mz=1
0x+(m-1)y+(1-m)z=0
0y+0y+(2-m-m²)z=-m-2

hallamos z en la tercera ecuación
(2-m-m²)z=-m-2
(m²+m-2)z=m+2

z=(m+2)/(m²+m-2)

hallamos y en la segunda ecuación
(m-1)y+(1-m)z=0
(m-1)y+(1-m)(m+2)/(m²+m-2)=0
(1-m)y=-(1-m)(m+2)/(m²+m-2)
y=-(m+2)/(m²+m-2)

hallamos x en la primera ecuación
x+y+mz=1
x-(m+2)/(m²+m-2)+m((m+2)/(m²+m-2))=1
x=(m+2)/(m²+m-2)-m((m+2)/(m²+m-2))
x=((m+2)-m(m+2))/(m²+m-2)
x=(m+2)(1-m)/(m²+m-2)

x=(m-m²+2-2m)/(m²+m-2)=(-m²-m+2)/(m²+m-2)=-(m²+m-2)/(m²+m-2)=-1
x=-1

y=-(m+2)/(m²+m-2)=-(m+2)/((m+2)(m-1))=-1/(m-1)=
 y=1/(1-m)    

z=(m+2)/(m²+m-2)=(m+2)/((m+2)(m-1))=1/(m-1)
z=1/(m-1)