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 = 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 (a2+1) (ab) (ac) (ab) (b2+1) (bc) (ca) (cb) (c2+1) = 1 a3+a 2b 2c ab2 b3+b 2c c2a 2b c3+c Applying R1 R1 + R2 + R3 = 1 a3+a+ 2+ 2 2b+b3+b+c2b 2c+b2c+c3+c ab2 b3+b 2c c2a 2b c3+c = 1 a( + + + ) ( + + + ) ( + + +1) ab2 b3+b 2c c2a 2b c3+c Taking (1+ 2+ 2+ 2) common from 1st Row = ( + + + ) a ab2 b3+b 2c c2a 2b c(c3+1) Taking a common from C1 ,b from C2 & c from C3 . = (1+ 2+ 2+ 2) 1 1 1 b2 b3+1 2 c2 2 c2+1 Applying C1 C1 C2 = (1+ 2+ 2+ 2) 1 1 b2 2 1 b2+1 2 c2 c2 2 c2+1 = (1+ 2+ 2+ 2) 1 1 1 b2+1 2 0 2 c2+1 Applying C2 C2 C3 = (1+ 2+ 2+ 2) 0 1 1 b2+1 2 2 0 2 2 1 c2+1 = (1+ 2+ 2+ 2) 0 1 1 1 2 0 1 c2+1 Expanding along R1 = (1+ 2+ 2+ 2) 1 1 2 1 2+1 0 1 2 0 2+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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