Consider the upper‑triangular matrix $D=\left(\begin{array}{ccc}\lambda & 2& 1\\ 0& \lambda -4& 5\\ 0& 0& \lambda +3\end{array}\right)$.\n(a) Find all real values of $\lambda$ for which $D$ is singular.\n(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.\n(c) If $D$ is non‑singular, express the solution $\mathbf{x}=(x,y,z{)}^T$ in terms of $\lambda$.

Step-by-step solution

1. For an upper‑triangular matrix, the determinant is the product of the diagonal entries: $\det D=\lambda (\lambda -4)(\lambda +3)$.
2. The matrix is singular when the determinant is zero, i.e. when any diagonal factor is zero:
3. • $\lambda =0$,
4. • $\lambda -4=0\Rightarrow \lambda =4$,
5. • $\lambda +3=0\Rightarrow \lambda =-3$.
6. Hence the singular values are $\lambda =0,4,-3$.
7. Now examine the system $D\mathbf{x}=\mathbf{c}$ with $\mathbf{c}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right)$ for each singular $\lambda$.
8. • For $\lambda =0$: the first equation becomes $0\cdot x+2y+z=1$, the second $-4y+5z=-2$, the third $3z=3\Rightarrow z=1$. Substituting $z=1$ into the second gives $-4y+5=-2\Rightarrow y=7/4$. The first equation then gives $2\cdot \frac{7}{4}+1=\frac{15}{2}\ne 1$, so the equations are contradictory – no solution.
9. • For $\lambda =4$: the second diagonal entry vanishes. The equations are $4x+2y+z=1$, $5z=-2$, $7z=3$. From the second $z=-2/5$, but the third demands $z=3/7$, contradiction – no solution.
10. • For $\lambda =-3$: the third diagonal entry is zero, giving $0\cdot z=3$, i.e. $0=3$, an immediate inconsistency – no solution.
11. Thus for every singular $\lambda$ the system is inconsistent.
12. When $\lambda$ is not $0$, $4$, or $-3$, $D$ is invertible. Solve by back‑substitution:
13. Third equation: $(\lambda +3)z=3\Rightarrow z=\frac{3}{\lambda +3}$.
14. Second equation: $(\lambda -4)y+5z=-2\Rightarrow y=\frac{-2-5z}{\lambda -4}$.
15. First equation: $\lambda x+2y+z=1\Rightarrow x=\frac{1-2y-z}{\lambda }$.
16. Substituting the expression for $z$ into the formula for $y$, and then both $z$ and $y$ into the formula for $x$ gives the solution in terms of $\lambda$ as stated.

Concept tested: Singularity of triangular matrix and solvability

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