Check sibling questions

Misc 4 - Show that either a + b + c = 0 or a = b = c - Making whole row/column one and simplifying

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


Transcript

Misc 4 If a, b and c are real numbers, and ∆ = b+c﷮c+a﷮a+b﷮c+a﷮a+b﷮b+c﷮a+b﷮b+c﷮c+a﷯﷯ = 0 , Show that either a + b + c = 0 or a = b = c Solving ∆ ∆ = b+c﷮c+a﷮a+b﷮c+a﷮a+b﷮b+c﷮a+b﷮b+c﷮c+a﷯﷯ Applying R1→ R1 + R2 + R3 = b+c+c+a+a+b﷮𝑐+𝑎+𝑎+𝑏+𝑏+𝑐﷮a+b+b+c+c+a﷮𝑐+𝑎﷮a+b﷮b+c﷮𝑎+𝑏﷮b+c﷮c+a﷯﷯ = 𝟐(𝐚+𝐛+𝐜)﷮𝟐(𝐚+𝐛+𝐜)﷮𝟐(𝐚+𝐛+𝐜)﷮𝑐+𝑎﷮a+b﷮b+c﷮𝑎+𝑏﷮b+c﷮c+a﷯﷯ Taking Common 2(𝑎+𝑏+𝑐) From R1 = 𝟐(𝐚+𝐛+𝐜) 1﷮1﷮1﷮𝑐+𝑎﷮a+b﷮b+c﷮𝑎+𝑏﷮b+c﷮c+a﷯﷯ Applying C2→ C2 – C1 = 2(a+b+c) 1﷮𝟏−𝟏﷮1﷮𝑐+𝑎﷮a+b−c−a﷮b+c﷮𝑎+𝑏﷮b+c−a−b﷮c+a﷯﷯ = 2(a+b+c) 1﷮𝟎﷮1﷮𝑐+𝑎﷮a−c﷮b+c﷮𝑎+𝑏﷮c−a﷮c+a﷯﷯ Applying C3→ C3 – C1 = 2(a+b+c) 1﷮0﷮𝟏−𝟏﷮𝑐+𝑎﷮b−c﷮b+c−c−a﷮𝑎+𝑏﷮c−a﷮c+a−a−b﷯﷯ = 2(a+b+c) 1﷮0﷮𝟎﷮𝑐+𝑎﷮b−c﷮b−a﷮𝑎+𝑏﷮c−a﷮c−b﷯﷯ Expanding determinant along R1 = 2(a+b+c) 1 𝑏−𝑐﷮𝑏−𝑎﷮𝑐−𝑎﷮𝑐−𝑏﷯﷯−0 𝑐+𝑎﷮𝑏−𝑎﷮𝑎+𝑏﷮𝑐−𝑏﷯﷯+0 𝑐+𝑎﷮𝑏−𝑐﷮𝑎+𝑏﷮𝑐−𝑎﷯﷯﷯ = 2(a+b+c) 1 𝑏−𝑐﷮𝑏−𝑎﷮𝑐−𝑎﷮𝑐−𝑏﷯﷯−0+0﷯ = 2(a+b+c) ((c – b) (b – c) – (b – a) (c – a)) = 2(a+b+c) ( – (c – b) (c – b) – (bc – ab – ac + a2)) = 2(a+b+c) ( – (c – b)2 – (bc – ab – ac + a2)) = 2(a+b+c) ( – (c2 + b2 – 2cd) – bc + ab + ac – a2) = 2(a+b+c) ( – c2 – b2 + 2cb – bc + ab + ac – a2) = 2(a+b+c) ( – (a2 + b2 + c2) + ab + bc + ac) ∴ ∆ = 2(a+b+c) ( – (a2 + b2 + c2) + ab + bc + ac) Given ∆ = 0 2(a+b+c) ( – (a2 + b2 + c2) + ab + bc + ac) = 0 So, either (a + b + c) = 0 or a = b = c Hence proved

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.