JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Matrices

The coefficient determinant of the system is
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& {\lambda }^2\\ 1& 3& \lambda \end{array}\right|$.
Expanding, $\Delta =(2\lambda -3{\lambda }^2)-(\lambda -{\lambda }^2)+(3-2)=-2{\lambda }^2+\lambda +1$. Setting $\Delta =0$ gives $2{\lambda }^2-\lambda -1=0$, i.e. $(2\lambda +1)(\lambda -1)=0$, so $\lambda =1$ or $\lambda =-\frac{1}{2}$.

For $\lambda =1$, the equations reduce to $x+y+z=5$, $x+2y+z=9$, $x+3y+z=\mu$. Subtracting successive pairs gives $y=4$ and $y=\mu -9$, so the system is consistent (with infinitely many solutions, since $\Delta =0$) exactly when $\mu =13$, and has no solution when $\mu \ne 13$. This confirms statements (1) and (2).

For $\lambda \ne 1$ and $\lambda \ne -\frac{1}{2}$, $\Delta \ne 0$ and the system has a unique solution for every $\mu$, in particular for $\mu =13$ — confirming statement (3).

However $\lambda =-\frac{1}{2}$ also satisfies $\lambda \ne 1$, and at this value $\Delta =0$ regardless of $\mu$; for a $\mu$ other than the single value that makes this particular system consistent, there is no solution at all — not a unique one. So $\lambda \ne 1,\mu \ne 13$ does not guarantee a unique solution, and statement (4) is the one that is NOT correct.