Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 17 / b

Collecting the coefficients in the given equation order yields the system of linear equations in matrix form

The determinant is $\det L=4-{\lambda }^2$, so there is exactly one solution whenever $\lambda \ne \pm 2$. 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 isFor $\lambda =2$ this reads $0=1$, and for $\lambda =-2$ it reads $0=5$. Both are inconsistent. Consequentlyand each permitted value has a unique solution.

For $\lambda \ne \pm 2$, the reduced equation gives $y=(3-\lambda )/(4-{\lambda }^2)$. Back substitution into $x=1-\lambda y$ and $z=1-4y$ gives all solutions:The nonzero determinant proves that there are no other solutions at these parameter values.