# Ex 4.2, 8

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 4.2, 8 By using properties of determinants, show that: (i) 1𝑎𝑎21𝑏𝑏21𝑐𝑐2 = (a - b) (b - c)(c – a) Taking L.H.S 1𝑎𝑎21𝑏𝑏21𝑐𝑐2 Applying R1 → R1 − R2 = 𝟏−𝟏𝑎−𝑏 𝑎2− 𝑏21𝑏𝑏21𝑐𝑐2 = 𝟎(𝑎−𝑏)(𝑎−𝑏)(𝑎+𝑏)1𝑏𝑏21𝑐𝑐2 = 0(𝐚−𝐛)(𝐚−𝐛) 𝐚−𝒃(a+b)1bb21cc2 Taking Common (a – b) from R1 = 𝐚−𝒃 01a+b1bb21cc2 Applying R2 → R2 − R3 = a−b 01a+b𝟏−𝟏b−cb2−c21cc2 = (a – b) 01a+𝑏𝟎b−c(b−c)(b+c)1cc2 Taking common (b – c) from R2 = (a – b) (b – c) 01a+b01b+c1cc2 Expanding Determinant along C1 = (a – b) (b – c) 0 1𝑏+𝑐𝑐𝑐2−0 1𝑎+𝑏𝑐𝑐2+1 1𝑎+𝑏1𝑏+𝑐 = (a – b) (b – c) 0−0+1 1𝑎+𝑏1𝑏+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) 111abca3b3c3 = (a – b) (b – c) (c – a) (a + b + c) Taking L.H.S 111abca3b3c3 Applying C1 → C1 − C2 = 𝟏−𝟏11a−bbc𝐚𝟑 −𝐛𝟑b3 c3 = 𝟎11a−bbc(𝐚 −𝐛)(𝐚𝟐+𝐛𝟐+𝐚𝐛) b3c3 = 011𝐚−𝐛bc(𝐚 −𝐛)(a2+b2+ab) b3c3 Taking Common (a – b) from C1 = (a – b) 0111bc(a2+b2+ab)b3c3 Applying C2 → C2 − C3 = (a – b) 0𝟏−𝟏11b−cc(a2+b2+ab)b3−c3c3 = (a – b) 0𝟎11b−cc(a2+b2+ab) b−c(b2+c2+bc)c3 Taking Common (b – c) from C2 = (a – b) (b – c) 00111ca2+b2+abb2+c2+bcc3 Expanding determinant along R1 = (a – b) (b – c) 0 1𝑐𝑏2+𝑐2+𝑏𝑐𝑐3−0 11𝑎2+𝑏2+𝑎𝑏𝑐3+1 11𝑎2+𝑏2+𝑎𝑏𝑏2+𝑐2+𝑏𝑐 = (a – b) (b – c) 0−0+1 11𝑎2+𝑏2+𝑎𝑏𝑏2+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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.