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Ex 4.2
Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 3 Deleted for CBSE Board 2023 Exams
Ex 4.2, 4 Deleted for CBSE Board 2023 Exams
Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams You are here
Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at Aug. 18, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 4.2, 8 By using properties of determinants, show that: (ii) |■8(1&1&[email protected]&b&[email protected]&b3&c3)| = (a – b) (b – c) (c – a) (a + b + c) Solving L.H.S |■8(1&1&[email protected]&b&[email protected]&b3&c3)| Applying C1 → C1 − C2 = |■8(𝟏−𝟏&1&[email protected]−b&b&[email protected]𝐚𝟑 −𝐛𝟑&b3 &c3)| = |■8(𝟎&1&[email protected]−b&b&[email protected](𝐚 −𝐛)(𝐚𝟐+𝐛𝟐+𝐚𝐛) &b3&c3)| = |■8(0&1&[email protected]𝐚−𝐛&b&[email protected](𝐚 −𝐛)(a2+b2+ab) &b3&c3)| Taking Common (a – b) from C1 = (a – b) |■8(0&1&[email protected]&b&[email protected](a2+b2+ab)&b3&c3)| Applying C2 → C2 − C3 = (a – b) |■8(0&𝟏−𝟏&[email protected]&b−c&[email protected](a2+b2+ab)&b3−c3&c3)| (x3 – y3 = (x – y)(x2 + y2 +xy)) = (a – b) |■8(0&𝟎&[email protected]&b−c&[email protected](a2+b2+ab)&(b−c)(b2+c2+bc)&c3)| Taking Common (b – c) from C2 = (a – b) (b – c) |■8(0&0&[email protected]&1&[email protected]+b2+ab&b2+c2+bc&c3)| Expanding determinant along R1 = (a – b) (b – c) (0|■8(1&𝑐@𝑏2+𝑐2+𝑏𝑐&𝑐3)|−0|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑐3)|+1|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑏2+𝑐2+𝑏𝑐)|) = (a – b) (b – c) (0−0+1|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑏2+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