JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Matrices

A linear system has a unique solution exactly when the determinant of its coefficient matrix is non-zero. Expanding the determinant $\left|\begin{array}{ccc}1+\alpha & \beta & 1\\ \alpha & 1+\beta & 1\\ \alpha & \beta & 2\end{array}\right|$, for instance by subtracting rows to clear the repeated $\alpha$ and $\beta$ columns, collapses it neatly to $\alpha +\beta +2$. A unique solution therefore needs $\alpha +\beta +2\ne 0$, i.e. $\alpha +\beta \ne -2$. Checking the pairs, $(2,4)$ gives $\alpha +\beta =6\ne -2$ and qualifies, whereas $(-3,1)$, $(-4,2)$ and $(1,-3)$ each sum to $-2$ and make the determinant vanish. The pair that yields a unique solution is $(2,4)$.