Mathematics - Determinant Question with Solution | TestHub

The given determinant vanishes, i.e., $$\left|\begin{array}{ccc}1& x-3& (x-3{)}^2\\ 1& x-4& (x-4{)}^2\\ 1& x-5& (x-5{)}^2\end{array}\right|=0$$ Expanding along ${\mathrm{C}}_1,$ we get $$\begin{array}{l}(x-4)(x-5{)}^2-(x-5)(x-4{)}^2-\{(x-3)(x-5{)}^2\\ -(x-5)(x-3{)}^2\}+(x-3)(x-4{)}^2\\ -(x-4)(x-3{)}^2=0\\ \Rightarrow (x-4)(x-5)(x-5-x+4)\\ -(x-3)(x-5)(x-5-x+3)\\ +(x-3)(x-4)(x-4-x+3)=0\end{array}$$ $$\Rightarrow -(x-4)(x-5)+2(x-3)(x-5)-(x-3)$$ $(x-4)=0$ $$\Rightarrow -x^2+9x-20+2x^2-16x+30-x^2+7x-12=0$$ $\Rightarrow -32+30=0\Rightarrow -2=0$ Which is not possible, hence no value of $\mathrm{x}$ satisfies the given condition.