Suppose D=a1 b1 c1a2 b2 c2a3 b3 c3 and D1=a1+pb1 b1+qc1 c1+ra1a2+pb2 b2+qc2 c2+ra2a3+pb3 b3+qc3 c3+ra3. Then

D′=a1+pb1b1+qc1c1+ra1a2+pb2b2+qc2c2+ra2a3+pb3b3+qc3c3+ra3D′=a1b1+qc1c1+ra1a2b2+qc2c2+ra2a3b3+qc3c3+ra3+pb1b1+qc1c1+ra1pb2b2+qc2c2+ra2pb3b3+qc3c3+ra3=a1 b1 c1a2 b2 c2a3 b3 c3+pb1qc1ra1pb2qc2ra2pb3qc3ra3=D+pqrb1 c1 a1b2 c2 a2b3 c3 a3D′=(1+pqr)D

Suppose D=a1 b1 c1a2 b2 c2a3 b3 c3 and D1=a1+pb1 b1+qc1 c1+ra1a2+pb2 b2+qc2 c2+ra2a3+pb3 b3+qc3 c3+ra3. Then

D′=a1+pb1b1+qc1c1+ra1a2+pb2b2+qc2c2+ra2a3+pb3b3+qc3c3+ra3D′=a1b1+qc1c1+ra1a2b2+qc2c2+ra2a3b3+qc3c3+ra3+pb1b1+qc1c1+ra1pb2b2+qc2c2+ra2pb3b3+qc3c3+ra3=a1 b1 c1a2 b2 c2a3 b3 c3+pb1qc1ra1pb2qc2ra2pb3qc3ra3=D+pqrb1 c1 a1b2 c2 a2b3 c3 a3D′=(1+pqr)D

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

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D’= D $(1 - pqr)$

D’= D

$D^{'} = D (1 + pqr)$

$D^{'} = D (1 + p + q + r)$

answer is D.

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Detailed Solution

$\begin{array}{l} D^{'} = |\begin{array}{lll} a_{1} + {pb}_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ a_{2} + {pb}_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ a_{3} + {pb}_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| \\ D^{'} = |\begin{array}{lll} a_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ a_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ a_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| + |\begin{array}{lll} {pb}_{1} & b_{1} + {qc}_{1} & c_{1} + {ra}_{1} \\ {pb}_{2} & b_{2} + {qc}_{2} & c_{2} + {ra}_{2} \\ {pb}_{3} & b_{3} + {qc}_{3} & c_{3} + {ra}_{3} \end{array}| \\ = |\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}| + |\begin{array}{ccc} p b_{1} & q c_{1} & r a_{1} \\ p b_{2} & q c_{2} & r a_{2} \\ p b_{3} & q c_{3} & r a_{3} \end{array}| \\ = D + p q r |\begin{array}{lll} b_{1} c_{1} a_{1} \\ b_{2} c_{2} a_{2} \\ b_{3} c_{3} a_{3} \end{array}| \\ D^{'} = (1 + pqr) D \end{array}$