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Last updated at Jan. 22, 2020 by Teachoo
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Ex 4.2, 8 By using properties of determinants, show that: (i) |โ 8(1&๐&๐2@1&๐&๐2@1&๐&๐2)| = (a - b) (b - c)(c โ a) Solving L.H.S |โ 8(1&๐&๐2@1&๐&๐2@1&๐&๐2)| Applying R1 โ R1 โ R2 = |โ 8(๐โ๐&๐โ๐&๐^2โ๐^2@1&๐&๐2@1&๐&๐2 ) | = |โ 8(๐&(๐โ๐)&(๐โ๐)(๐+๐)@1&๐&๐2@1&๐&๐2 ) | = |โ 8(0(๐โ๐)&(๐โ๐)&(๐โ๐)(a+b)@1&b&b2@1&c&c2 ) | Taking Common (a โ b) from R1 = (๐โ๐) |โ 8(0&1&a+b@1&b&b2@1&c&c2 ) | Applying R2 โ R2 โ R3 = (aโb) |โ 8(0&1&a+b@๐โ๐&bโc&b2โc2@1&c&c2 ) | = (a โ b) |โ 8(0&1&a+๐@๐&bโc&(bโc)(b+c)@1&c&c2 ) | Taking common (b โ c) from R2 = (a โ b) (b โ c) |โ 8(0&1&a+b@0&1&b+c@1&c&c2 ) | Expanding Determinant along C1 = (a โ b) (b โ c) ( 0|โ 8(1&๐+๐@๐&๐2)|โ0|โ 8(1&๐+๐@๐&๐2)|+1|โ 8(1&๐+๐@1&๐+๐)|) = (a โ b) (b โ c) ( 0โ0+1|โ 8(1&๐+๐@1&๐+๐)|) = (a โ b) (b โ c) (1(b + c) โ 1(a + b) ) = (a โ b) (b โ c) (b + c โ a โ b) = (a โ b) (b โ c)(c โ a) = R.H.S Hence Proved Ex 4.2, 8 By using properties of determinants, show that: (ii) |โ 8(1&1&1@a&b&c@a3&b3&c3)| = (a โ b) (b โ c) (c โ a) (a + b + c) Solving L.H.S |โ 8(1&1&1@a&b&c@a3&b3&c3)| Applying C1 โ C1 โ C2 = |โ 8(๐โ๐&1&1@aโb&b&c@๐๐ โ๐๐&b3 &c3)| = |โ 8(๐&1&1@aโb&b&c@(๐ โ๐)(๐๐+๐๐+๐๐) &b3&c3)| = |โ 8(0&1&1@๐โ๐&b&c@(๐ โ๐)(a2+b2+ab) &b3&c3)| Taking Common (a โ b) from C1 = (a โ b) |โ 8(0&1&1@1&b&c@(a2+b2+ab)&b3&c3)| Applying C2 โ C2 โ C3 = (a โ b) |โ 8(0&๐โ๐&1@1&bโc&c@(a2+b2+ab)&b3โc3&c3)| (x3 โ y3 = (x โ y)(x2 + y2 +xy)) = (a โ b) |โ 8(0&๐&1@1&bโc&c@(a2+b2+ab)&(bโc)(b2+c2+bc)&c3)| Taking Common (b โ c) from C2 = (a โ b) (b โ c) |โ 8(0&0&1@1&1&c@a2+b2+ab&b2+c2+bc&c3)| Expanding determinant along R1 = (a โ b) (b โ c) (0|โ 8(1&๐@๐2+๐2+๐๐&๐3)|โ0|โ 8(1&1@๐2+๐2+๐๐&๐3)|+1|โ 8(1&1@๐2+๐2+๐๐&๐2+๐2+๐๐)|) = (a โ b) (b โ c) (0โ0+1|โ 8(1&1@๐2+๐2+๐๐&๐2+๐2+๐๐)|) = (a โ b) (b โ c) (1((b2 + c2 + bc) โ (a2 + b2 + ab)) = (a โ b) (b โ c) (b2 + c2 + bc โ a2 โ b2 โ ab) = (a โ b) (b โ c) (c2 โ a2 + bc โ ab) = (a โ b) (b โ c) ((c โ a) (c + a) + b (c โ a)) = (a โ b) (b โ c) ((c โ a) (c + a + b)) = (a โ b) (b โ c) ((c โ a) (a + b + c)) = R.H.S Hence Proved
Ex 4.2
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