If the system of linear equations
`{:(x+2ay+az=0),(x+3by+bz=0),(x+4cy+cz=0):}`
has a non zero solutions, then `a, b, c` are in

If the system of linear equations x + 2ay + az = 0 x + 3by + bz = 0 x + 4cy + cz = 0 has a non-zero solution, then a, b, c

If the system of linear equations x+2ay +az =0,x+3by+bz =0 and x+4cy + cz =0 has a non-zero solution, then a, b, c

If the system of equations x + 4ay + az = 0 and x + 3by + bz = 0 and x + 2cy + cz = 0 have a nonzero solution, then a, b, c ( != 0) are in

If the system of linear equations x+a y+a z=0,\ x+b y+b z=0,\ x+c y+c z=0 has a non zero solution then (a) System is always non trivial solutions (b) System is consistent only when a=b=c (c) If a!=b!=c then x=0,\ y=t ,\ z=-t\ AAt in R (d) If a=b=c\ t h e n\ y=t_1, z=t_2,\ x=-a(t_1+t_2)AAt_1, t_2 in R

If the system of linear equations x+ky+3z=0 3x+ky-2z=0 2x+4y-3z=0 has a non-zero solution (x,y,z) then (xz)/(y^2) is equal to

if the system of equations {:(ax+y+2z=0),(x+2y+z=b),(2x+y+az=0):} has no solution then (a+b) can be equals to :

Prove that the system of equations in xa +y+z=0 , x +by +z=0 , x +y+cz=0 has a non - trivial solution then 1/(1-a) + 1/(1-b)+ 1/(1-c)=

If the system of linear equations x=cy+bz , y=az+cx , z=bx+ay has a non trivial solution show that a^2+b^2+c^2+2abc=1

If the system of equations bx + ay = c, cx + az = b, cy + bz = a has a unique solution, then

If the system of linear equations ax +by+ cz =0 cx + ay + bz =0 bx + cy +az =0 where a,b,c, in R are non-zero and distinct, has a non-zero solution, then: