A 3×3 real matrix A has characteristic polynomial λ^3-6λ^2+11λ-6. The value of |adj(A)+2I| is

The characteristic polynomial factors as $(\lambda-1)(\lambda-2)(\lambda-3)$, so the eigenvalues of $A$ are $1,2,3$ and $|A|=6$. Since $\operatorname{adj}A=|A|\,A^{-1}$, the eigenvalues of $\operatorname{adj}A$ are $\dfrac61,\dfrac62,\dfrac63=6,3,2$. Adding $2I$ shifts every eigenvalue by $2$, giving $8,5,4$. The determinant is the product of the eigenvalues: $8\cdot5\cdot4=160$. One may check with the diagonal matrix $\operatorname{diag}(1,2,3)$, whose adjugate is $\operatorname{diag}(6,3,2)$. The eigenvalue route needs no explicit $A$ and is the shortest path to the determinant of the shifted adjugate.