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Ex 4.2, 14 - Using properties |a2+1 ab| = 1 + a2 + b2 + c2

Ex 4.2, 14 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 14 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.2, 14 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.2, 14 - Chapter 4 Class 12 Determinants - Part 5

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Transcript

Ex 4.2, 14 By using properties of determinants, show that: |β– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |β– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying & Dividing by abc = 𝒂𝒃𝒄/𝒂𝒃𝒄 |β– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 β†’ aR1 , R2 β†’ bR3 , R3 β†’ bR3 ) = 1/π‘Žπ‘π‘ |β– 8(𝒂(a2+1)&𝒂(ab)&𝒂(ac)@𝒃(ab)&𝐛(b2+1)&𝒃(bc)@𝐜(ca)&𝒄(cb)&𝐜(c2+1))| = 1/π‘Žπ‘π‘ |β– 8(a3+a&π‘Ž2b&π‘Ž2c@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| Applying R1 β†’ R1 + R2 + R3 = 1/π‘Žπ‘π‘ |β– 8(a3+a+π‘Žπ‘2+𝑐2π‘Ž&π‘Ž2b+b3+b+c2b&π‘Ž2c+b2c+c3+c@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| = 1/π‘Žπ‘π‘ |β– 8(a(𝐚𝟐+𝟏+π’ƒπŸ+π’„πŸ)&𝑏(π’‚πŸ+π›πŸ+𝟏+𝐜𝟐)&𝑐(π’‚πŸ+π›πŸ+𝐜𝟐"+1" )@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| Taking (1+π‘Ž2+𝑏2+𝑐2) common from 1st Row = ((𝟏 + π’‚πŸ + π’ƒπŸ + π’„πŸ))/π‘Žπ‘π‘ |β– 8(a&𝑏&𝑐@ab2&b3+b&𝑏2c@c2a&𝑐2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = 𝒂𝒃𝒄 ( (1 + π‘Ž2 + 𝑏2 + 𝑐2))/π‘Žπ‘π‘ |β– 8(1&1&1@b2&b3+1&𝑏2@c2&𝑐2&c2+1)| Applying C1 β†’ C1 βˆ’ C2 = (1+π‘Ž2+𝑏2+𝑐2) |β– 8(πŸβˆ’πŸ&1&1@b2βˆ’π‘2βˆ’1&b2+1&𝑏2@c2βˆ’c2&𝑐2&c2+1)| = (1+π‘Ž2+𝑏2+𝑐2) |β– 8(𝟎&1&1@βˆ’1&b2+1&𝑏2@0&𝑐2&c2+1)| Applying C2 β†’ C2 βˆ’ C3 = (1+π‘Ž2+𝑏2+𝑐2) |β– 8(0&πŸβˆ’πŸ&1@βˆ’1&b2+1βˆ’π‘2&𝑏2@0&𝑐2βˆ’π‘2βˆ’1&c2+1)| = (1+π‘Ž2+𝑏2+𝑐2) |β– 8(0&𝟎&1@βˆ’1&1&𝑏2@0&βˆ’1&c2+1)| Expanding along R1 = (1+π‘Ž2+𝑏2+𝑐2)(1|β– 8(1&𝑏2@βˆ’1&𝑐2+1)|" – 0" |β– 8(βˆ’1&𝐢2@0&𝑐2+1)|" + 0" |β– 8(1&1@0&βˆ’1)|) = (1 + a2 + b1 + c2) (0 – 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.