A) Divisible by k
B) Divisible by \[{{k}^{2}}\]
C) Divisible by \[{{k}^{3}}\]
D) None of these
Correct Answer: A
Solution :
[a] Since, \[{{A}_{1}}{{B}_{1}}{{C}_{1}},\,{{A}_{2}}{{B}_{2}}{{C}_{2}}\] and \[{{A}_{3}}{{B}_{3}}{{C}_{3}}\] are divisible by k, therefore; \[100{{A}_{1}}+10{{B}_{1}}+{{C}_{1}}={{n}_{1}}k;\] \[100{{A}_{2}}+10{{B}_{2}}+{{C}_{2}}\] \[={{n}_{2}}k;\,\,100{{A}_{3}}+10{{B}_{3}}+{{C}_{3}}={{n}_{3}}k\] (where \[{{n}_{1}},{{n}_{2}},{{n}_{3}}\] are integers) \[Now,\,\,\,\,\Delta =\left| \begin{matrix} {{A}_{1}} & {{B}_{1}} & {{C}_{1}} \\ {{A}_{2}} & {{B}_{2}} & {{C}_{2}} \\ {{A}_{3}} & {{B}_{3}} & {{C}_{3}} \\ \end{matrix} \right|\] \[=\,\,\,\left| \begin{matrix} {{A}_{1}} & {{B}_{1}} & 100{{A}_{1}}+10{{B}_{1}}+{{C}_{1}} \\ {{A}_{2}} & {{B}_{2}} & 100{{A}_{2}}+10{{B}_{2}}+{{C}_{2}} \\ {{A}_{3}} & {{B}_{3}} & 100{{A}_{3}}+10{{B}_{2}}+{{C}_{3}} \\ \end{matrix} \right|\] [Applying \[{{C}_{3}}\to {{C}_{3}}+10{{C}_{2}}+100{{C}_{1}}\]] \[=\left| \begin{matrix} {{A}_{1}} & {{B}_{1}} & {{n}_{1}}k \\ {{A}_{2}} & {{B}_{2}} & {{n}_{2}}k \\ {{A}_{3}} & {{B}_{3}} & {{n}_{3}}k \\ \end{matrix} \right|=k\left| \begin{matrix} {{A}_{1}} & {{B}_{1}} & {{n}_{1}} \\ {{A}_{2}} & {{B}_{2}} & {{n}_{2}} \\ {{A}_{3}} & {{B}_{3}} & {{n}_{3}} \\ \end{matrix} \right|=k{{\Delta }_{1}}\] \[\Rightarrow \Delta \] is divisible by k [Since, elements of \[{{\Delta }_{1}}\] are integers \[\therefore \,{{\Delta }_{1}}\] is an integer.]
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