A) \[\Delta '=\Delta \]
B) \[\Delta '=\Delta \,\,(1-pqr)\]
C) \[\Delta '=\Delta \,\,(1+p+q+r)\]
D) \[\Delta '=\Delta \,\,(1+pqr)\]
Correct Answer: D
Solution :
[d] \[\Delta '=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} p{{b}_{1}} & q{{c}_{1}} & r{{a}_{1}} \\ p{{b}_{2}} & q{{c}_{2}} & r{{a}_{2}} \\ p{{b}_{3}} & q{{c}_{3}} & r{{a}_{3}} \\ \end{matrix} \right|\] [All other determinants vanish] \[=\Delta +pqr\Delta =\Delta (1+pqr)\]You need to login to perform this action.
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