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.