Consider the $4\times 4$ matrix $E(\lambda )=\left[\begin{array}{cccc}\lambda & 1& 1& 1\\ 1& \lambda & 1& 1\\ 1& 1& \lambda & 1\\ 1& 1& 1& \lambda \end{array}\right].$\n\n(a) Find $\det E(\lambda )$ as a polynomial in $\lambda$.\n(b) Determine all real values of $\lambda$ for which $E(\lambda )$ is singular.

Step-by-step solution

1. Step 1: Observe that each off‑diagonal entry equals $1$. Write $E(\lambda )=(\lambda -1)I_4+J$, where $I_4$ is the $4\times 4$ identity matrix and $J$ is the matrix of all ones.
2. Step 2: The matrix $J$ has rank $1$ with eigenvalues $4$ (once) and $0$ (thrice). Therefore the eigenvalues of $E(\lambda )$ are $(\lambda -1)+4=\lambda +3$ (once) and $(\lambda -1)+0=\lambda -1$ (thrice).
3. Step 3: The determinant of a matrix equals the product of its eigenvalues. Hence $\det E(\lambda )=(\lambda +3)(\lambda -1{)}^3.$
4. Step 4: For part (b), a matrix is singular when its determinant is zero. Set $(\lambda +3)(\lambda -1{)}^3=0$.
5. Step 5: Solve each factor: $\lambda +3=0\Rightarrow \lambda =-3$, and $\lambda -1=0\Rightarrow \lambda =1$ (with multiplicity three).
6. Step 6: Conclude that $E(\lambda )$ is singular exactly for $\lambda =1$ or $\lambda =-3$.

Concept tested: Determinant of a matrix with a parameter (using eigenvalue method or row operations)

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked