Mathematics - Determinant Question with Solution | TestHub

For the system to have no solution, the determinant of the coefficient matrix must be zero, i.e., $\Delta =0$.$$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\lambda & 1\\ 1& 1& 1+\lambda \end{array}\right|$$ Apply row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$: $$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right|=1\cdot (\lambda \cdot \lambda -0)={\lambda }^2$$ Setting $\Delta =0$ gives ${\lambda }^2=0$, so $\lambda =0$. Now, check the system for $\lambda =0$:$x+y+z=1$ $x+y+z=2$ $x+y+z=3$ Subtracting the first equation from the second yields $0=1$, which is a contradiction. Thus, the system has no solution when $\lambda =0$.