Ex 4.2, 14 - Using properties |a2+1 ab| = 1 + a2 + b2 + c2 - Using Property 5 (Determinant as sum of two or more determinants)

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Ex 4.2, 14 By using properties of determinants, show that: a2+1﷮ab﷮ac﷮ab﷮b2+1﷮bc﷮ca﷮cb﷮c2+1﷯﷯ = 1 + a2 + b2 + c2 Taking L.H.S a2+1﷮ab﷮ac﷮ab﷮b2+1﷮bc﷮ca﷮cb﷮c2+1﷯﷯ Multiplying & Dividing by abc = 𝒂𝒃𝒄﷮𝒂𝒃𝒄﷯ 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 → aR1 , R2 → bR3 , R3 → bR3 ) = 1﷮𝑎𝑏𝑐﷯ 𝒂(a2+1)﷮𝒂(ab)﷮𝒂(ac)﷮𝒃(ab)﷮𝐛(b2+1)﷮𝒃(bc)﷮𝐜(ca)﷮𝒄(cb)﷮𝐜(c2+1)﷯﷯ = 1﷮𝑎𝑏𝑐﷯ a3+a﷮𝑎2b﷮𝑎2c﷮ab2﷮b3+b﷮𝑏2c﷮c2a﷮𝑐2b﷮c3+c﷯﷯ Applying R1 → R1 + R2 + R3 = 1﷮𝑎𝑏𝑐﷯ a3+a+𝑎𝑏2+𝑐2𝑎﷮𝑎2b+b3+b+c2b﷮𝑎2c+b2c+c3+c﷮ab2﷮b3+b﷮𝑏2c﷮c2a﷮𝑐2b﷮c3+c﷯﷯ = 1﷮𝑎𝑏𝑐﷯ a(𝐚𝟐+𝟏+𝒃𝟐+𝒄𝟐)﷮𝑏(𝒂𝟐+𝐛𝟐+𝟏+𝐜𝟐)﷮𝑐(𝒂𝟐+𝐛𝟐+𝐜𝟐+1)﷮ab2﷮b3+b﷮𝑏2c﷮c2a﷮𝑐2b﷮c3+c﷯﷯ Taking (1+𝑎2+𝑏2+𝑐2) common from 1st Row = (𝟏+𝒂𝟐+𝒃𝟐+𝒄𝟐)﷮𝑎𝑏𝑐﷯ 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+𝑎2+𝑏2+𝑐2)﷮𝑎𝑏𝑐﷯ 1﷮1﷮1﷮b2﷮b3+1﷮𝑏2﷮c2﷮𝑐2﷮c2+1﷯﷯ Applying C1 → C1 − C2 = (1+𝑎2+𝑏2+𝑐2) 𝟏−𝟏﷮1﷮1﷮b2−𝑏2−1﷮b2+1﷮𝑏2﷮c2−c2﷮𝑐2﷮c2+1﷯﷯ = (1+𝑎2+𝑏2+𝑐2) 𝟎﷮1﷮1﷮−1﷮b2+1﷮𝑏2﷮0﷮𝑐2﷮c2+1﷯﷯ Applying C2 → C2 − C3 = (1+𝑎2+𝑏2+𝑐2) 0﷮𝟏−𝟏﷮1﷮−1﷮b2+1−𝑐2﷮𝑏2﷮0﷮𝑐2−𝑐2−1﷮c2+1﷯﷯ = (1+𝑎2+𝑏2+𝑐2) 0﷮𝟎﷮1﷮−1﷮1﷮𝑏2﷮0﷮−1﷮c2+1﷯﷯ Expanding along R1 = (1+𝑎2+𝑏2+𝑐2) 1 1﷮𝑏2﷮−1﷮𝑐2+1﷯﷯ – 0 −1﷮𝐶2﷮0﷮𝑐2+1﷯﷯ + 0 1﷮1﷮0﷮−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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