# Misc 14 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 14 Using properties of determinants, prove that: 11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q = 1 Taking L.H.S 11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q Applying R2→ R2 – 2R1 = 11+p1+p+q𝟐−𝟐(𝟏)3+2p−2(1+p)4+3𝑝+2𝑝−2(1+𝑝+𝑞)36+3p−210+6p+3q = 11+p1+p+q𝟎12+𝑝36+3p10+6p+3q Applying R3→ R3 – 3R2 = 11+p1+p+q012+p𝟑−𝟑(𝟏)6+3𝑝−3(1+𝑝)10+6𝑝+3𝑞−3(1+𝑝+𝑞) = 11+p1+p+q012+p𝟎37+3𝑝 Expanding determinant along C1 = 1 12+p37+3p – 0 1+𝑝2+p37+3p + 0 1+𝑝1+p+q12+p = 1 12+p37+3p – 0 + 0 = 1 (7 + 3p – 3 (2 + p)) = 7 + 3p – 6 – 3p = 1 = R.H.S Hence Proved

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.