question_answer

The value of the determinant\[\left| \begin{matrix}    kb & {{k}^{^{2}}}+{{a}^{2}} & 1  \\    kb & {{k}^{2}}+{{b}^{2}} & 1  \\    kc & {{k}^{2}}+{{c}^{2}} & 1  \\ \end{matrix} \right|\]is

A)  \[k(a+b)(b+c)(c+a)\]

B)  \[k\,abc({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]

C)  \[k(a-b)(b-c)(c-a)\]

D)  \[k(a+b-c)(b+c-a)(c+a-b)\]

Correct Answer: C

Solution :

[c] we have, \[=k(a-b)(b-c)(c-a)\]