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

Step-by-step solution

1. The determinant of an upper‑triangular matrix equals the product of its diagonal entries.
2. Compute $\det (F)=\lambda (\lambda -3)(\lambda +2)$.
3. Set the determinant to zero: $\lambda (\lambda -3)(\lambda +2)=0$, giving $\lambda =0$, $\lambda =3$, or $\lambda =-2$.
4. Write the three equations represented by $F\mathbf{x}=\mathbf{b}$:
5. 1) $\lambda x_1+2x_2+x_3=3$
6. 2) $(\lambda -3)x_2+4x_3=-1$
7. 3) $(\lambda +2)x_3=2$.
8. Analyse each singular $\lambda$:
9. • $\lambda =0$: Equation (3) gives $2x_3=2\Rightarrow x_3=1$. Equation (2): $-3x_2+4(1)=-1\Rightarrow -3x_2=-5\Rightarrow x_2=5/3$. Equation (1): $2x_2+x_3=2\cdot 5/3+1=13/3\ne 3$ → contradiction.
10. • $\lambda =3$: Equation (3): $5x_3=2\Rightarrow x_3=2/5$. Equation (2): $0\cdot x_2+4(2/5)=-1\Rightarrow 8/5=-1$, impossible → contradiction.
11. • $\lambda =-2$: Equation (3): $0\cdot x_3=2$, which is false → contradiction.
12. In each case the augmented matrix gains a higher rank than the coefficient matrix, so the system is inconsistent.
13. Therefore, for $\lambda =0,3,-2$ the matrix is singular and the given system has no solution.

Concept tested: Singular matrices with a parameter and consistency

Published 11 September 2026 · How this solution was prepared and checked