Consider the upper‑triangular matrix $D=\left(\begin{array}{ccc}\lambda & 2& 1\\ 0& \lambda -5& 4\\ 0& 0& \lambda +2\end{array}\right)$. (a) Find all real values of $\lambda$ for which $D$ is singular. (b) For each such $\lambda$, discuss whether the system $D\mathbf{x}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right)$ has a solution.

Step-by-step solution

1. For an upper‑triangular matrix, the determinant is the product of the diagonal entries: $\det D=\lambda (\lambda -5)(\lambda +2)$.
2. Set the determinant to zero: $\lambda (\lambda -5)(\lambda +2)=0$, giving $\lambda =0$, $\lambda =5$, or $\lambda =-2$. These are the singular values.
3. Now examine the system for each singular $\lambda$ using back‑substitution:
4. \textbf{(i) $\lambda =0$:} $D$ becomes $\left(\begin{array}{ccc}0& 2& 1\\ 0& -5& 4\\ 0& 0& 2\end{array}\right)$. Row 3 gives $2z=3\Rightarrow z=\frac{3}{2}$. Row 2 gives $-5y+4z=-2\Rightarrow -5y+6=-2\Rightarrow y=\frac{8}{5}$. Row 1 requires $2y+z=1$, but $2\cdot \frac{8}{5}+\frac{3}{2}=\frac{47}{10}\ne 1$. Hence inconsistency.
5. \textbf{(ii) $\lambda =5$:} $D$ becomes $\left(\begin{array}{ccc}5& 2& 1\\ 0& 0& 4\\ 0& 0& 7\end{array}\right)$. Row 3 gives $7z=3\Rightarrow z=\frac{3}{7}$. Row 2 gives $4z=-2\Rightarrow z=-\frac{1}{2}$, which contradicts the previous value; the system is inconsistent.
6. \textbf{(iii) $\lambda =-2$:} $D$ becomes $\left(\begin{array}{ccc}-2& 2& 1\\ 0& -7& 4\\ 0& 0& 0\end{array}\right)$. Row 3 reduces to $0=3$, an immediate contradiction. Hence no solution.
7. Since each singular case leads to a contradiction, the system $D\mathbf{x}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right)$ has no solution for $\lambda =0,5,-2$.

Concept tested: Singularity of parameter matrix and consistency of linear system

Published 19 September 2026 · How this solution was prepared and checked