The system will have a unique solution for any value of lambda for
which the 3x3 determinant of the coefficients of the left hand side
is not 0.
In your case all lambda but 1 or -2
1 1 lambda
1 lambda 1
lambda 1 1
This matrix has a determinant:
lambda + lambda + lambda - 1 -1 - lambda3
This has to be different from 0, to make the matrix non-singular, and thus guarantee that it has only one solution.
This third-order polynomial has the following solutions:
-2 and 1
Any other lambda will give you a system that has a unique solution.
You can find one solution and then divide the polynomial.
For example, it is easy to see that 1 is a solution.
3l - l^3 - 2 = 0
This means that 3l - l^3 - 2 = p(l ) * (l - 1)
where p(l ) is a second order polynomial. You can find p(l ) by dividing:
(3l - l^3 - 2) / (l -1) = -l^2 - l + 2
and using the solving formula for 2nd order polynomials
p(l ) = 0 <=> l = -2 V l = 1