• # question_answer Let $Y=SX$and $Z=tX$ such that$\left| \begin{matrix} X & Y & Z \\ {{X}_{1}} & {{Y}_{1}} & {{Z}_{1}} \\ {{X}_{2}} & {{Y}_{2}} & {{Z}_{2}} \\ \end{matrix} \right|+\left| \begin{matrix} {{S}_{1}} & {{t}_{1}} \\ {{S}_{2}} & {{t}_{2}} \\ \end{matrix} \right|={{X}^{n}},$all the variables being differentiable functions of x and lower suffices denote the derivative with respect to x. Then n = A) $1$                       B)        $2$                       C) $3$        D)        $4$

[c] $\Delta =\left| \begin{matrix} X & SX & tX \\ {{X}_{1}} & S{{X}_{1}}+{{S}_{1}}X & t{{X}_{1}}+{{t}_{1}}X \\ {{X}_{2}} & S{{X}_{2}}+2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & t{{X}_{2}}+2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X \\ \end{matrix} \right|$ $=\left| \begin{matrix} X & 0 & 0 \\ {{X}_{1}} & {{S}_{1}}X & {{t}_{1}}X \\ {{X}_{2}} & 2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & 2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X \\ \end{matrix} \right|$ (Applying ${{C}_{2}}\to {{C}_{2}}-S{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-t{{C}_{1}}$) $={{X}^{2}}\left| \begin{matrix} {{S}_{1}} & {{t}_{1}} \\ 2{{S}_{1}}{{X}_{1}}+{{S}_{2}}X & 2{{t}_{1}}{{X}_{1}}+{{t}_{2}}X \\ \end{matrix} \right|$ $={{X}^{3}}\left| \begin{matrix} {{S}_{1}} & {{t}_{1}} \\ {{S}_{2}} & {{t}_{2}} \\ \end{matrix} \right|$ (Applying ${{R}_{2}}\to {{R}_{2}}-2{{X}_{1}}{{R}_{1}}$ )              $\therefore \,\,\,\,\,\,n=3$