Ex 4.2, 8 - Class 12

Transcript

Ex 4.2, 8 By using properties of determinants, show that: (i) 1﷮𝑎﷮𝑎2﷮1﷮𝑏﷮𝑏2﷮1﷮𝑐﷮𝑐2﷯﷯ = (a - b) (b - c)(c – a) Taking L.H.S 1﷮𝑎﷮𝑎2﷮1﷮𝑏﷮𝑏2﷮1﷮𝑐﷮𝑐2﷯﷯ Applying R1 → R1 − R2 = 𝟏−𝟏﷮𝑎−𝑏﷮ 𝑎﷮2﷯− 𝑏﷮2﷯﷮1﷮𝑏﷮𝑏2﷮1﷮𝑐﷮𝑐2 ﷯ ﷯ = 𝟎﷮(𝑎−𝑏)﷮(𝑎−𝑏)(𝑎+𝑏)﷮1﷮𝑏﷮𝑏2﷮1﷮𝑐﷮𝑐2 ﷯ ﷯ = 0(𝐚−𝐛)﷮(𝐚−𝐛)﷮ 𝐚−𝒃﷯(a+b)﷮1﷮b﷮b2﷮1﷮c﷮c2 ﷯ ﷯ Taking Common (a – b) from R1 = 𝐚−𝒃﷯ 0﷮1﷮a+b﷮1﷮b﷮b2﷮1﷮c﷮c2 ﷯ ﷯ Applying R2 → R2 − R3 = a−b﷯ 0﷮1﷮a+b﷮𝟏−𝟏﷮b−c﷮b2−c2﷮1﷮c﷮c2 ﷯ ﷯ = (a – b) 0﷮1﷮a+𝑏﷮𝟎﷮b−c﷮(b−c)(b+c)﷮1﷮c﷮c2 ﷯ ﷯ Taking common (b – c) from R2 = (a – b) (b – c) 0﷮1﷮a+b﷮0﷮1﷮b+c﷮1﷮c﷮c2 ﷯ ﷯ Expanding Determinant along C1 = (a – b) (b – c) 0 1﷮𝑏+𝑐﷮𝑐﷮𝑐2﷯﷯−0 1﷮𝑎+𝑏﷮𝑐﷮𝑐2﷯﷯+1 1﷮𝑎+𝑏﷮1﷮𝑏+𝑐﷯﷯﷯ = (a – b) (b – c) 0−0+1 1﷮𝑎+𝑏﷮1﷮𝑏+𝑐﷯﷯﷯ = (a – b) (b – c) (1(b + c) – 1(a + b) ) = (a – b) (b – c) (b + c – a – b) = (a – b) (b – c)(c – a) = R.H.S Hence Proved Ex 4.2, 8 By using properties of determinants, show that: (ii) 1﷮1﷮1﷮a﷮b﷮c﷮a3﷮b3﷮c3﷯﷯ = (a – b) (b – c) (c – a) (a + b + c) Taking L.H.S 1﷮1﷮1﷮a﷮b﷮c﷮a3﷮b3﷮c3﷯﷯ Applying C1 → C1 − C2 = 𝟏−𝟏﷮1﷮1﷮a−b﷮b﷮c﷮𝐚𝟑 −𝐛𝟑﷮b3 ﷮c3﷯﷯ = 𝟎﷮1﷮1﷮a−b﷮b﷮c﷮(𝐚 −𝐛)(𝐚𝟐+𝐛𝟐+𝐚𝐛) ﷮b3﷮c3﷯﷯ = 0﷮1﷮1﷮𝐚−𝐛﷮b﷮c﷮(𝐚 −𝐛)(a2+b2+ab) ﷮b3﷮c3﷯﷯ Taking Common (a – b) from C1 = (a – b) 0﷮1﷮1﷮1﷮b﷮c﷮(a2+b2+ab)﷮b3﷮c3﷯﷯ Applying C2 → C2 − C3 = (a – b) 0﷮𝟏−𝟏﷮1﷮1﷮b−c﷮c﷮(a2+b2+ab)﷮b3−c3﷮c3﷯﷯ = (a – b) 0﷮𝟎﷮1﷮1﷮b−c﷮c﷮(a2+b2+ab)﷮ b−c﷯(b2+c2+bc)﷮c3﷯﷯ Taking Common (b – c) from C2 = (a – b) (b – c) 0﷮0﷮1﷮1﷮1﷮c﷮a2+b2+ab﷮b2+c2+bc﷮c3﷯﷯ Expanding determinant along R1 = (a – b) (b – c) 0 1﷮𝑐﷮𝑏2+𝑐2+𝑏𝑐﷮𝑐3﷯﷯−0 1﷮1﷮𝑎2+𝑏2+𝑎𝑏﷮𝑐3﷯﷯+1 1﷮1﷮𝑎2+𝑏2+𝑎𝑏﷮𝑏2+𝑐2+𝑏𝑐﷯﷯﷯ = (a – b) (b – c) 0−0+1 1﷮1﷮𝑎2+𝑏2+𝑎𝑏﷮𝑏2+𝑐2+𝑏𝑐﷯﷯﷯ = (a – b) (b – c) (1((b2 + c2 + bc) – (a2 + b2 + ab)) = (a – b) (b – c) (b2 + c2 + bc – a2 – b2 – ab) = (a – b) (b – c) (c2 – a2 + bc – ab) = (a – b) (b – c) ((c – a) (c + a) + b (c – a)) = (a – b) (b – c) ((c – a) (c + a + b)) = (a – b) (b – c) ((c – a) (a + b + c)) = R.H.S Hence Proved