For which real values of $k$ is the 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 the solvability of $E\mathbf{x}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$.

Step-by-step solution

1. Note that $E$ is an upper triangular matrix; its determinant is the product of the diagonal entries: $\det E=k(k-1)(k-2)$.
2. The matrix is singular when its determinant is zero, i.e., when $k=0$, $k=1$, or $k=2$.
3. Case $k=0$: $E$ becomes $\left(\begin{array}{ccc}0& 1& 2\\ 0& -1& 3\\ 0& 0& -2\end{array}\right)$. The first equation reads $0\cdot x_1+x_2+2x_3=1$, the second $-x_2+3x_3=2$, the third $-2x_3=3$. From the third equation $x_3=-\frac{3}{2}$, then the second gives $-x_2+3(-\frac{3}{2})=2\Rightarrow -x_2-\frac{9}{2}=2\Rightarrow x_2=-\frac{13}{2}$. Substituting in the first equation $-\frac{13}{2}+2(-\frac{3}{2})=1$ leads to $-\frac{13}{2}-3=1$, which is false. Hence the system is inconsistent; no solution for $k=0$.
4. Case $k=1$: $E$ becomes $\left(\begin{array}{ccc}1& 1& 2\\ 0& 0& 3\\ 0& 0& -1\end{array}\right)$. The third equation gives $-x_3=3\Rightarrow x_3=-3$. The second equation becomes $0\cdot x_2+3(-3)=2\Rightarrow -9=2$, which is false, so we made an algebra error. Re‑evaluate: for $k=1$, the second diagonal entry $k-1=0$, so the second row is $(0,0,3)$, giving $3x_3=2\Rightarrow x_3=\frac{2}{3}$. The third row is $(0,0,-1)$, giving $-x_3=3\Rightarrow x_3=-3$. The two values of $x_3$ contradict, so the system is inconsistent; no solution for $k=1$ as well. (Correct analysis shows inconsistency for $k=1$).
5. Case $k=2$: $E$ becomes $\left(\begin{array}{ccc}2& 1& 2\\ 0& 1& 3\\ 0& 0& 0\end{array}\right)$. The third equation $0\cdot x_1+0\cdot x_2+0\cdot x_3=3$ is impossible, so the system is inconsistent; no solution for $k=2$.
6. Hence the only singular values are $k=0,1,2$, and for each the augmented matrix has rank greater than the coefficient matrix, giving no solution. (If a value yielded equal ranks, it would give infinitely many solutions; none does here).
7. Therefore the matrix is singular for $k=0,1,2$; the given system has no solution for all these values.

Concept tested: Determinant of triangular matrix and solvability of singular systems

Published 6 September 2026 · How this solution was prepared and checked