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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

Miscellaneous

Misc 1

Misc. 2 Important

Misc 3

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 Important

Misc 12 Important

Misc. 13

Misc 14 You are here

Misc. 15 Important

Misc. 16 Important

Misc 17 Important

Misc 18

Misc 19 Important

Matrices and Determinants - Formula Sheet and Summary

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.