1. Chapter 4 Class 12 Determinants
2. Serial order wise

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Example 15 If x, y, z are different and Δ = x﷮x2﷮1+x3﷮y﷮y2﷮1+y3﷮z﷮z2﷮1+z3﷯﷯ = 0 , then show that 1 + xyz = 0 Solving ∆ = x﷮x2﷮1+x3﷮y﷮y2﷮1+y3﷮z﷮z2﷮1+z3﷯﷯ Here, expanding elements of C3 into two determinants = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + x﷮x2﷮x3﷮y﷮y2﷮y3﷮z﷮z2﷮z3﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + x﷮x2﷮x3﷮y﷮y2﷮y3﷮z﷮z2﷮z3﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯+ xyz 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ = (−1) x﷮1﷮x2﷮y﷮1﷮y2﷮z﷮1﷮z2﷯﷯ + xyz 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ = (−1)(−1) 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ + xyz 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ = 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ + xyz 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ = 1﷮x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷯﷯ (1 + xyz) Using R2 → R2 – R1 and R3 → R3 – R1 = 1﷮x﷮x2﷮𝟏 −𝟏﷮y−x﷮y2 −x2﷮𝟏−𝟏﷮z−x﷮z2 −x2﷯﷯ (1+ xyz) = 1﷮x﷮x2﷮𝟎﷮(y−x)﷮ y −x﷯(y+x)﷮𝟎﷮(z−x)﷮ z −x﷯(z+x)﷯﷯ (1+ xyz) Taking common factor (y – x) from R2 & (z – x) from R3 = (1 + xyz) (y – x) (z – x) 1﷮x﷮x2﷮0﷮1﷮y+x﷮0﷮1﷮z+x﷯﷯ Expanding determinant = (1 + xyz) (y – x) (z – x) (z – y) 1 1﷮𝑦+𝑥﷮1﷮𝑧+𝑥﷯﷯ – 0 𝑥﷮𝑥2﷮1﷮𝑧+𝑥﷯﷯ + 0 𝑥﷮𝑥2﷮1﷮𝑦+𝑥﷯﷯﷯ = (1 + xyz) (y – x) (z – x) (1 (y + x) – (y + x) + 0 + 0) = (1 + xyz) (y – x) (z – x) (z + y – y – x) = (1 + xyz) (y – x) (z – x) (z – y) ∴ ∆ = (1 + xyz) (y – x) (z – x) (z – y) Since ∆ = 0 given (1 + xyz) (y – x) (z – x) (z – y) = 0 Since it is given that x, y, z all are different, i.e., y – x ≠ 0, z – x ≠ 0, z – y ≠ 0, So, only Possibility is (1 + xyz) = 0 Hence Proved