Consider the upper‑triangular matrix D=\begin{pmatrix}\lambda&2&1\0&\lambda-5&4\0&0&\lambda+2\end{\pmatrix}. (a) Which values of $\lambda$ make $D$ singular? (b) For each such value, does the system Dmathbf{x}=\begin{pmatrix}1\-2\3\end{\pmatrix} have a solution? (c) If $D$ is nonsingular, express the solution $\mathbf{x}$ in terms of $\lambda$. Choose the correct answer from the options below.

Step-by-step solution

1. Because $D$ is upper‑triangular, its determinant is the product of the diagonal entries: $detD=lambda(lambda-5)(lambda+2)$.
2. Set $detD=0$ to find singularity: $lambda=0$, $lambda=5$, or $lambda=-2$.
3. These are the only real values of $lambda$ that make $D$ singular.
4. For each singular $lambda$, $detD=0$ implies the system has either no solution or infinitely many solutions; a detailed check is required for each case.
5. If $\lambda$ is none of $0,5,-2$, $D$ is nonsingular. Then solve the system by back‑substitution: $z=\frac{3}{\lambda +2}$, $y=\frac{-2-4z}{\lambda -5}$, $x=\frac{1-2y-z}{\lambda }$.
6. This gives $\mathbf{x}$ in terms of $lambda$ for nonsingular cases.

Concept tested: Determinant of upper‑triangular matrix and solvability of linear system

Published 13 October 2026 · How this solution was prepared and checked