Consider the parameter 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 $\lambda$, determine whether the system $D\mathbf{x}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right)$ has a solution. If it has, describe the solution set.

Step-by-step solution

1. Since $D$ is upper‑triangular, $\det (D)=\lambda (\lambda -5)(\lambda +2)$.
2. Set determinant to zero to find singular values: $\lambda =0,5,-2$.
3. For each value, solve the system or check consistency:
4. •$\lambda =0:$ $D=\left(\begin{array}{ccc}0& 2& 1\\ 0& -5& 4\\ 0& 0& 2\end{array}\right)$.\nThird equation gives $2z=3\Rightarrow z=\frac{3}{2}$.\nSecond gives $-5y+4z=-2\Rightarrow y=\frac{8}{5}$.\nFirst gives $2y+z=1$, which fails for the obtained $y,z$; hence inconsistent—no solution.
5. •$\lambda =5:$ $D=\left(\begin{array}{ccc}5& 2& 1\\ 0& 0& 4\\ 0& 0& 7\end{array}\right)$.\nSecond gives $4z=-2\Rightarrow z=-\frac{1}{2}$.\nThird gives $7z=3\Rightarrow z=\frac{3}{7}$; contradiction—no solution.
6. •$\lambda =-2:$ $D=\left(\begin{array}{ccc}-2& 2& 1\\ 0& -7& 4\\ 0& 0& 0\end{array}\right)$.\nThird row is zero; system reduces to two equations in $x,y,z$.\nSolve: from second, $-7y+4z=-2\Rightarrow y=\frac{4z+2}{7}$.\nSubstitute into first: $-2x+2y+z=1$.\nReplace $y$ and simplify to $-2x+\frac{8z+4}{7}+z=1$.\nAfter clearing denominators, obtain $14x=15z-3\to x=\frac{15z-3}{14}$.\nParameter $z$ is free; thus infinite solutions: $\mathbf{x}=\left(\begin{array}{c}\frac{15z-3}{14}\frac{4z+2}{7}z\end{array}\right)$.

Concept tested: Singular upper‑triangular matrix and solution existence

Published 17 September 2026 · How this solution was prepared and checked