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

Step-by-step solution

1. For an upper‑triangular matrix, the determinant is the product of diagonal entries: $\det C=\lambda (\lambda -2)(\lambda +1)$.
2. Set the determinant to zero: $\lambda (\lambda -2)(\lambda +1)=0$ giving $\lambda =0$, $\lambda =2$, $\lambda =-1$.
3. Case $\lambda =0$: $C$ becomes $\left(\begin{array}{ccc}0& 1& 4\\ 0& -2& 5\\ 0& 0& 1\end{array}\right)$. Write the three equations:
4. (1) $x_2+4x_3=2$, (2) $-2x_2+5x_3=-3$, (3) $x_3=4$.
5. From (3) $x_3=4$. Substitute into (1): $x_2+16=2\Rightarrow x_2=-14$. Substitute $x_2$ and $x_3$ into (2): $-2(-14)+5\cdot 4=28+20=48\ne -3$. Hence contradiction → no solution.
6. Case $\lambda =2$: $C$ becomes $\left(\begin{array}{ccc}2& 1& 4\\ 0& 0& 5\\ 0& 0& 3\end{array}\right)$. Equations:
7. (1) $2x_1+x_2+4x_3=2$, (2) $5x_3=-3$, (3) $3x_3=4$.
8. From (2) $x_3=-3/5$, but (3) gives $x_3=4/3$, a conflict. Hence no solution.
9. Case $\lambda =-1$: $C$ becomes $\left(\begin{array}{ccc}-1& 1& 4\\ 0& -3& 5\\ 0& 0& 0\end{array}\right)$. Equations:
10. (1) $-x_1+x_2+4x_3=2$, (2) $-3x_2+5x_3=-3$, (3) $0=4$.
11. Equation (3) is impossible, so the system is inconsistent.
12. Thus for every singular value of $\lambda$, the system has no solution.

Concept tested: Singular matrix and solvability of linear system

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