 Solution

For the Cholesky decomposition $A=L{L}^T$, choose $L$ lower triangular with positive diagonal whenever $A$ is positive-definite. Successive entries areThe $(3,2)$ equation gives $l_{31}{l}_{21}+l_{32}{l}_{22}=2+3\lambda$, so $l_{32}=3\sqrt{\lambda +2}$. Finally,Thus for $\lambda >-2$,It is nonsingular, and for every nonzero vector $v$, $v^{T}Av=|L^{T}v{|}^2>0$.

Necessity also follows from the leading principal determinants $2$, $2(\lambda +2)$ and $6(\lambda +2)$, or directly from the second elimination pivot. HenceAt $\lambda =-2$ the displayed limiting factor still satisfies $A=L{L}^T$, but has rank two and zero second diagonal entry: $A$ is positive semidefinite, not positive-definite. For $\lambda <-2$, taking $v=(2,1,0{)}^T$ gives $v^{T}Av=\lambda +2<0$, so no real factor $L{L}^T$ exists.