# Example 15

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 15 If x, y, z are different and Δ = xx21+x3yy21+y3zz21+z3 = 0 , then show that 1 + xyz = 0 Solving ∆ = xx21+x3yy21+y3zz21+z3 Here, expanding elements of C3 into two determinants = xx21yy21zz21 + xx2x3yy2y3zz2z3 = xx21yy21zz21 + xx2x3yy2y3zz2z3 = xx21yy21zz21+ xyz 1xx21yy21zz2 = (−1) x1x2y1y2z1z2 + xyz 1xx21yy21zz2 = (−1)(−1) 1xx21yy21zz2 + xyz 1xx21yy21zz2 = 1xx21yy21zz2 + xyz 1xx21yy21zz2 = 1xx21yy21zz2 (1 + xyz) Using R2 → R2 – R1 and R3 → R3 – R1 = 1xx2𝟏 −𝟏y−xy2 −x2𝟏−𝟏z−xz2 −x2 (1+ xyz) = 1xx2𝟎(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) 1xx201y+x01z+x Expanding determinant = (1 + xyz) (y – x) (z – x) (z – y) 1 1𝑦+𝑥1𝑧+𝑥 – 0 𝑥𝑥21𝑧+𝑥 + 0 𝑥𝑥21𝑦+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

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Chapter 4 Class 12 Determinants

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .