and
. Then
A) D'=D
B) D'=D(1-pqr)
C) D'=D(1+p+q+r)
D) D'=D(1+pqr)
Correct Answer: D
Solution :
[d]
In the first determinant, apply \[{{C}_{3}}\to {{C}_{3}}-r{{C}_{1}}\]and then\[{{C}_{2}}\to {{C}_{2}}-q{{C}_{3}}\]. In the second determinant, take p common form \[{{C}_{1}}\]and then apply \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1.}}\]Then take q common form \[{{C}_{2}}\]and apply \[{{C}_{3}}\to {{C}_{3}}-{{C}_{2}}\]. Finally taking r common form \[{{C}_{3}}\]we have ultimately\[D'=(1+pqr)D\].
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