1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Ex 4.2

Transcript

Ex 4.2, 9 By using properties of determinants, show that: x x2 yz y y2 zx z z2 xy = (x y) (y z) (z x)(xy + yz + zx) Taking L.H.S x x2 yz y y2 zx z z2 xy Applying R1 R1 R2 = x x2 y2 yz yz y y2 zx z z2 xy = ( + ) z ( ) y y2 zx z z2 xy Taking out (x y) common from R1 = (x y) 1 + z y y2 zx z z2 xy Applying R2 R2 R3 = (x y) 1 + z y z y2 z2 zx xy z z2 xy = (x y) 1 + z ( )( + ) x( ) z z2 xy Taking out (y z) common from R2 = (x y) (y z) 1 + z 1 + x z z2 xy Applying R1 R1 R2 = (x y) (y z) 1 + ( + ) z ( x) 1 + x z z2 xy = (x y) (y z) + z+x 1 + x z z2 xy = (x y) (y z) 0 x z 1 + x z z2 xy = (x y) (y z) 0 ( ) ( ) 1 + x z z2 xy Taking out (z x) from R1 = (x y) (y z) (z x) 0 1 1 1 + x z z2 xy Applying C2 C2 C3 = (x y) (y z) (z x) 0 1 ( 1) 1 1 + ( ) x z z2 xy xy = (x y) (y z) (z x) 0 1+1 1 1 + + x z z2 xy xy = (x y) (y z) (z x) 0 1 1 + + x z z2 xy xy Expanding Determinant along R1 = (x y) (y z) (z x) 0 + + 2 0 1 + 1 1 + + 2 = (x y) (y z) (z x) 0 0+ 1 1 + + 2 = (x y) (y z) (z x) (0 0 1((z2 xy) z (x + y + z))) = (x y) (y z) (z x) ( (z2 xy) + z (x + y + z)) = (x y) (y z) (z x) ( z2 + xy + + zx + zy + z2) = (x y) (y z) (z x) ( xy + zx + zy + z2 z2) = (x y) (y z) (z x) ( xy + yz + zx) = R.H.S Hence proved

Ex 4.2