Define $F(\lambda )=\left[\begin{array}{ccc}\lambda & 2& 3\\ 2& \lambda & 4\\ 3& 4& \lambda \end{array}\right]$.\n\n(a) Find $\det F(\lambda )$ as a polynomial in $\lambda$.\n(b) Determine all real values of $\lambda$ for which $F(\lambda )$ is singular.\n(c) Let $G=\mathrm{adj}F(1)$. Compute $\det G$.

Step-by-step solution

1. Compute the determinant using cofactor expansion or the rule for $3\times 3$ matrices: \n$\det F(\lambda )=\lambda ({\lambda }^2-4\cdot 4)-2(2\lambda -4\cdot 3)+3(2\cdot 4-\lambda \cdot 3)$.
2. Simplify each term: $\lambda ({\lambda }^2-16)={\lambda }^3-16\lambda$, $-2(2\lambda -12)=-4\lambda +24$, $3(8-3\lambda )=24-9\lambda$.
3. Add the three results: ${\lambda }^3-16\lambda -4\lambda +24+24-9\lambda ={\lambda }^3-29\lambda +48$.
4. Thus $\det F(\lambda )={\lambda }^3-29\lambda +48$.
5. To find singular values, solve ${\lambda }^3-29\lambda +48=0$. By trying integer factors of 48 none satisfy, so the cubic has three real irrational roots. Using numerical methods (e.g., Newton–Raphson) we obtain approximations: ${\lambda }_1\approx 1.90$, ${\lambda }_2\approx 4.20$, ${\lambda }_3\approx -6.10$.
6. For part (c), first evaluate $\det F(1)$. Substitute $\lambda =1$ into the polynomial: $1^3-29\cdot 1+48=20$. Hence $\det F(1)=20$.
7. Recall the relation $\det (\mathrm{adj}A)=(\det A{)}^{n-1}$ for an $n\times n$ matrix. Here $n=3$, so $\det G=(\det F(1){)}^2=20^2=400$.
8. Therefore the required determinant of $G$ is $400$.

Concept tested: Determinant of a parameter matrix and determinant of an adjoint

Published 15 September 2026 · How this solution was prepared and checked