JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Determinants

Each entry $(i+j\alpha {)}^2$ expands as $i^2+2ij\alpha +j^2{\alpha }^2$, so every column is a combination of the three fixed vectors $a=(1,4,9{)}^T$, $b=(1,2,3{)}^T$ and $e=(1,1,1{)}^T$: column $j$ equals $a+2j\alpha b+j^2{\alpha }^{2}e$. Because the determinant is multilinear and alternating, only the terms that take $a$, $b$ and $e$ in three different columns survive; gathering them gives the determinant as $4{\alpha }^3\det [abe]$. Since $\det [abe]=-2$, the determinant equals $-8{\alpha }^3$. Setting $-8{\alpha }^3=-648\alpha$ and cancelling the non-zero $\alpha$ gives $8{\alpha }^2=648$, so ${\alpha }^2=81$ and $\alpha =\pm 9$.