JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Matrices

The coefficient determinant is $D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& \alpha \\ 1& 3& 5\end{array}\right|=6-2\alpha$. A unique solution exists precisely when $D\ne 0$, that is when $\alpha \ne 3$; for $\alpha =-3$ the determinant is non-zero, so a unique solution exists. When $\alpha =3$ the determinant vanishes. Subtracting the first equation from the second gives $y+2z=4$, while subtracting the second from the third gives $y+2z=\beta -10$; these agree only if $\beta =14$. Hence $\alpha =3,\beta =14$ yields infinitely many solutions, whereas $\alpha =3,\beta \ne 14$ (for instance $\beta =24$) yields no solution. Therefore the assertion that the system has a unique solution for $\alpha =3,\beta \ne 14$ is false, because in that case the system is inconsistent.