# Misc 6 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Miscellaneous

Misc 1

Misc. 2 Important Deleted for CBSE Board 2022 Exams

Misc 3

Misc 4 Deleted for CBSE Board 2022 Exams

Misc 5

Misc 6 Important Deleted for CBSE Board 2022 Exams You are here

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 Important Deleted for CBSE Board 2022 Exams

Misc 12 Important Deleted for CBSE Board 2022 Exams

Misc. 13 Deleted for CBSE Board 2022 Exams

Misc 14 Deleted for CBSE Board 2022 Exams

Misc. 15 Important Deleted for CBSE Board 2022 Exams

Misc. 16 Important

Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Misc 18 (MCQ)

Misc 19 (MCQ) Important

Matrices and Determinants - Formula Sheet and Summary Important

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

Misc 6 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