For which real values of $k$ is the upper‑triangular matrix $E=\left(\begin{array}{ccc}k& 1& 2\\ 0& k-1& 3\\ 0& 0& k-2\end{array}\right)$ singular? For each such $k$, discuss whether the system $E\mathbf{x}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$ has a solution.

Step-by-step solution

1. A triangular matrix is singular iff its determinant is zero. Compute $\det E=k(k-1)(k-2)$.
2. Set $k(k-1)(k-2)=0$ to find singular values: $k=0,1,2$. Actually check: determinant uses diagonal entries: $k$, $k-1$, $k-2$; product zero gives $k=0$ or $k=1$ or $k=2$.
3. Correct the matrix: the third diagonal element is $k-2$, so singular values are $k=0,1,2$.
4. Now solve $E\mathbf{x}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$. Write equations:$kx+y+2z=1$, $(k-1)y+3z=2$, $(k-2)z=3$.
5. For each singular value:
6. • If $k=0$: third equation becomes $-2z=3\to z=-1.5$. Substituting into second gives $-1y+3(-1.5)=2\to -y-4.5=2\to y=-6.5$. First equation: $0\cdot x-6.5+2(-1.5)=1\to -6.5-3=1\to -9.5=1$ contradiction. Thus no solution.
7. • If $k=1$: third equation becomes $-1z=3\to z=-3$. Second equation: $0y+3(-3)=2\to -9=2$ contradiction. No solution.
8. • If $k=2$: third equation gives $0z=3$ → impossible. No solution.
9. Hence for all singular values the system has no solution. (The earlier statement of infinitely many solutions for a value is incorrect.)

Concept tested: Singular triangular matrix and solvability of linear system

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked