Home
Class 12
MATHS
If x@ + y^2 + z^2!=0, x=cy + bz, y = az ...

If `x@ + y^2 + z^2!=0, x=cy + bz, y = az + cx`, and `z = bx + ay`, then `a^2 + b^2 + c^2 + 2abc=`

Text Solution

Verified by Experts

Given systems of equations can be rewritten as -x +cy +by =0, cx -y + az =0 and bx+ ay-z =0
Above system of equations are homogeneous equation. Since x,y and z are not all zero, so it has non-trivial solution.
Therefore, the coefficient of determinant must be zero.
`therefore |{:(-1, c, b),(c, -1, a),(b, a, -1):}| =0`
`rArr -1(1-a^(2)) -c(-c-ab) + b(ca+b) =0`
`rArr a^(2) + b^(2) +c^(2) +2abc-1 =0`
`rArr a^(2) + b^(2) + c^(3) + 2abc =1`
Promotional Banner

Similar Questions

Explore conceptually related problems

If the system of equations x= cy +bz, y=az +cx and z=bx+ay has a non - trivial solution then

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

Prove that |a x-b y-c z a y+b x c x+a z a y+b x b y-c z-a x b z+c y c x+a z b z+c y c z-a x-b y|=(x^2+y^2+z^2)(a^2+b^2+c^2)(a x+b y+c z)dot

If the system of linear equation x + 2ay + az = 0, x + 3by + bz = 0, x + 4cy + cz = 0 has a non-trival solution then show that a, b, c are in H.P.

If a x1 2+b y1 2+c z1 2=a x2 2+b y2 2+c z2 2=a x3 2+b y3 2+c z3 2=d ,a x2 3+b y_2y_3+c z_2z_3=a x_3x_1+b y_3y_1+c z_3z_1=a x_1x_2+b y_1y_2+c z_1z_2=f, then prove that |x_1y_1z_1x_2y_2z_2x_3y_3z_3|=(d-f){((d+2f))/(a b c)}^(1//2)

Given that the system of equations x=cy+bz ,y=az+cx , z=bx +ay has nonzero solutions and and at least one of the a,b,c is a proper fraction. a^(2)+b^(2)+c^(2) is

Given that the system of equations x=cy+bz ,y=az+cx , z=bx +ay has nonzero solutions and and at least one of the a,b,c is a proper fraction. abc is

Given that the system of equations x=cy+bz ,y=az+cx , z=bx +ay has nonzero solutions and and at least one of the a,b,c is a proper fraction. System has solution such that

If lines x=ay+b, z= cy +d " and " x=a' z+b y=c' z+ d' are perpendicular then