JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Matrices

For a non-trivial solution the determinant of the coefficient matrix must vanish: $\left|\begin{array}{ccc}1& -c& -c\\ c& -1& c\\ c& c& -1\end{array}\right|=0$. Expanding along the first row, $1(1-c^2)-c(c+c^2)-c(c^2+c)=1-3c^2-2c^3$, so the condition is $2c^3+3c^2-1=0$. This factorises as $(c+1{)}^2(2c-1)=0$, giving $c=-1$ as a repeated root or $c=\frac{1}{2}$. The greatest of these values is $c=\frac{1}{2}$.