Misc 6 - Prove that determinant = 4a2b2c2 - Class 12 NCERT

Misc 6 - Chapter 4 Class 12 Determinants - Part 2
Misc 6 - Chapter 4 Class 12 Determinants - Part 3 Misc 6 - Chapter 4 Class 12 Determinants - Part 4 Misc 6 - Chapter 4 Class 12 Determinants - Part 5

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Question 4 Prove that |■8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = 4a2b2c2 Solving L.H.S |■8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = |■8(𝐚(𝑎)&𝐛(c)&𝐜(a+c)@𝐚(𝑎+𝑏)&𝐛(b)&𝐜(a)@𝐚(𝑏)&𝐛(b+c)&𝐜(c))| Taking out a common from C1 ,b common from C2 & c common from C3 = abc |■8(a&c&a+c@𝑎+𝑏&b&a@𝑏&b+c&c)| Applying C1→ C1 + C2 + C3 = abc |■8(a+𝑐+𝑎+𝑐&c&a+c@𝑎+𝑏+𝑏+𝑎&b&a@𝑏+𝑏+𝑐+𝑐&b+c&c)| = abc |■8(2a+2𝑐&c&a+c@2a+2𝑏&b&a@2𝑏+2𝑐&b+c&c)| = abc |■8(𝟐(𝑎+𝑐)&c&a+c@𝟐(𝑎+𝑏)&b&a@𝟐(𝑏+𝑐)&b+c&c)| Taking out 2 Common from C2 = 2a𝑏𝑐|■8((𝑎+𝑐)&c&a+c@(𝑎+𝑏)&b&a@(𝑏+𝑐)&b+c&c)| Applying C3→ C3 – C1 = 2a𝑏𝑐|■8((𝑎+𝑐)&c&a+c−(𝐚+𝐜)@(𝑎+𝑏)&b&a−(a+b)@(𝑏+𝑐)&b+c&c−(b+c))| = 2a𝑏𝑐|■8(𝑎+𝑐&c&𝟎@𝑎+𝑏&b&−𝑏@𝑏+𝑐&b+c&−b)| Applying C2→ C2 – C1 = 2abc |■8(𝑎+𝑐&c−(𝑎+𝑐)&0@𝑎+𝑏&b−(𝑎+𝑏)&−b@𝑏+𝑐&b+c−(𝑏+𝑐)&−b)| = 2abc |■8(𝑎+𝑐&−𝒂&0@𝑎+𝑏&−𝒂&−𝐛@𝑏+𝑐&0&−𝐛)| Taking out –a common from C2, –b common from C3 = 2abc (–a) (–b) |■8(𝑎+𝑐&1&0@𝑎+𝑏&1&1@𝑏+𝑐&0&1)| Expanding Determinant along R1 = 2a2b2c ((a+c)|■8(1&1@0&1)|−1|■8(𝑎+𝑏&1@𝑏+𝑐&1)|+0|■8(𝑎+𝑏&1@𝑏+𝑐&0)|) = 2a2b2c ((a+c)|■8(1&1@0&1)|−1|■8(𝑎+𝑏&1@𝑏+𝑐&1)|+0) = 2a2b2c ((a + c) (1 – 0) – 1(a + b – (b + c))) = 2a2b2c ((a + c) (1) – 1 (a + b – b – c)) = 2a2b2c (a + c – a + c) = 2a2b2c (2c) = 4a2b2c = R.H.S Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.