A solutions set of the equations x+2y+z=1,x+3y+4z=k,x+5y+10z=`k^2`
`(5 lambda-1,1-3 lambda,lambda)`
`(1+5 lambda,-3 lambda,lambda)`
`(1+6 lambda,-2 lambda,lambda)`
` (1-6"lambda,lambda,lambda)`

p(lambda)^4+q(lambda)^3+r(lambda)^2+s(lambda)+t = |((lambda^2 + 3lambda) , lambda -1 , lambda+3) , (lambda +1 , -2lambda , lambda-4 ) ,(lambda-3 , lambda+4 , 3lambda)| find t=?

If |[lambda^(2)+3 lambda,lambda-1,lambda+3],[lambda+1,2-lambda,lambda-4],[lambda-3,lambda+4,3 lambda]|=p lambda^(4)+q lambda^(3)+r lambda^(2)+s lambda+t then t=

If plambda^4+qlambda^3+rlambda^2+slambda+t=|(lambda^2+3lambda,lambda-1,lambda+3),(lambda+1,2-lambda,lambda-4),(lambda-3,lambda+4,3lambda)|, then value of t is

If the matrix A = [[lambda_(1)^(2), lambda_(1)lambda_(2), lambda_(1) lambda_(3)],[lambda_(2)lambda_(1),lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]] is idempotent, the value of lambda_(1)^(2) + lambda_(2)^(2) + lambda _(3)^(2) is

If the two roots of the equation (lambda-2)(x^2+x+1)^2-(lambda+2)(x^4+x^2+1) are real and equal for lambda=lambda_1,lambda_2 then lambda_1+lambda_2=0 (b) |lambda_1-lambda_2|=6 (c) lambda_1+lambda_2=32 (d) all of these

If the points A(lambda, 2lambda), B(3lambda,3lambda) and C(3,1) are collinear, then lambda=

Find the solution of the system of equations (a^(2)+lambda)x+aby+acz=lambda,abx+(b^(2)+lambda)y+bcz=lambda,(b^(2)+lambda)y+bcz=lambda,acx+bcy+(c^(2)+lambda)z=lambda where a+b+c=0, and lambda!=0

If |[lambda^(2),-lambda,2 lambda-1],[lambda+2,1-lambda,lambda],[lambda+2,lambda+1,-lambda]|=a lambda^(4)+b lambda^(3)+c lambda^(2)+d lambda+e Then values of a, b, c, d and e are.

The system of equations lambda x+(lambda+1)y+(lambda-1)z=0,(lambda+1)x+lambda y+(lambda+z)z=0,(lambda-1)x+(lambda+2)y+lambda z= bas a non-trivial solutions for

A system of equations lambda x+y+z=1,x+lambda y+z=lambda,x+y+lambda z=lambda^(2) have no solution then value of lambda is