= Solution
The <determinant> is $\det L=4-\lambda^2$, so there is exactly one solution whenever $\lambda\ne\pm2$. A zero <determinant> does not alone distinguish inconsistency from infinitely many solutions, so check those values separately.
Eliminate $x$ and $z$ using $x=1-\lambda y$ and $z=1-4y$. The remaining equation is
$$
(4-\lambda^2)y=3-\lambda.
$$
For $\lambda=2$ this reads $0=1$, and for $\lambda=-2$ it reads $0=5$. Both are inconsistent. Consequently
$$
\boxed{\text{Solutions exist exactly for }\lambda\in\mathbb R\setminus\{-2,2\},}
$$
and each permitted value has a unique solution.
Back to article page
