Example 15 - Chapter 4 Class 12 Determinants

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Example 15 If x, y, z are different and = 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 y x y2 x2 z x z2 x2 (1+ xyz) = 1 x x2 (y x) y x (y+x) (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 + 1 + 0 2 1 + + 0 2 1 + = (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