Suppose ∆=a1 b1 c1a2 b2 c2a3 b3 c3 and ∆'=a1+pb1 b1+qc1 c1+ra1a2+pb2 b2+qc2 c2+ra2a3+pb3 b3+qc3 c3+ra3. Then

∆'=a1+pb1b1+qc1c1+ra1a2+pb2b2+qc2c2+ra2a3+pb3b3+qc3c3+ra3=a1b1+qc1c1+ra1a2b2+qc2c2+ra2a3b3+qc3c3+ra3+pb1b1+qc1c1+ra1pb2b2+qc2c2+ra2pb3b3+qc3c3+ra3In the first determinant, apply C3→C3−rC1 and then C2→C2−qC3.In the second determinant, take p common from C1 and then applyC2→C2−C1. Then take q common from C2 and applyC3→C3−C2. Finally taking r common from C3, we havefinally D' = (1 + pqr)D.

Suppose ∆=a1 b1 c1a2 b2 c2a3 b3 c3 and ∆'=a1+pb1 b1+qc1 c1+ra1a2+pb2 b2+qc2 c2+ra2a3+pb3 b3+qc3 c3+ra3. Then

∆'=a1+pb1b1+qc1c1+ra1a2+pb2b2+qc2c2+ra2a3+pb3b3+qc3c3+ra3=a1b1+qc1c1+ra1a2b2+qc2c2+ra2a3b3+qc3c3+ra3+pb1b1+qc1c1+ra1pb2b2+qc2c2+ra2pb3b3+qc3c3+ra3In the first determinant, apply C3→C3−rC1 and then C2→C2−qC3.In the second determinant, take p common from C1 and then applyC2→C2−C1. Then take q common from C2 and applyC3→C3−C2. Finally taking r common from C3, we havefinally D' = (1 + pqr)D.

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Suppose $∆ = |\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}|$ and $∆^{'} = |\begin{array}{lll} a_{1} + {pb}_{1} b_{1} + {qc}_{1} c_{1} + {ra}_{1} \\ a_{2} + {pb}_{2} b_{2} + {qc}_{2} c_{2} + {ra}_{2} \\ a_{3} + {pb}_{3} b_{3} + {qc}_{3} c_{3} + {ra}_{3} \end{array}|$ . Then

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D' = D

D' = D(1 - pqr)

D' = D(1 + p + q + r)

D' = D(1 + pqr)

answer is D.

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Detailed Solution

$\begin{matrix} ∆' = |\begin{array}{lll} a_{1} + {pb}_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ a_{2} + {pb}_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ a_{3} + {pb}_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| \\ = |\begin{array}{lll} a_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ a_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ a_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| + |\begin{array}{lll} {pb}_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ {pb}_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ {pb}_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| \end{matrix}$