# Misc 11 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 11 Using properties of determinants, prove that: 𝛼a2β+𝛾ββ2𝛾+𝛼𝛾𝛾2𝛼+β = (β – 𝛾) (𝛾 – 𝛼) (𝛼 – β) (a + β + 𝛾) Taking L.H.S 𝛼𝛼2β+yββ2y+𝛼yy2𝛼+β Applying C1→ C1 + C3 = 𝜶+𝜷+𝜸𝛼2β+𝛾𝛃+𝜸+𝜶β2𝛾+𝛼𝜸+𝜶+𝜷𝛾2𝛼+β Taking (α + β + 𝛾) common from C1 = (α + β + 𝜸) 1𝛼2β+𝛾1β2𝛾+𝛼1𝛾2𝛼+β Applying R2→ R2 – R1 = (α + β + 𝛾) 1a2β+𝛾𝟏−𝟏β2−a2𝛾+𝛼−𝛽−𝛾1y2𝛼+𝛽 = (α + β + 𝛾) 1a2β+𝛾𝟎(β−a)(𝛽+𝛼)−(𝛽−𝛼)1y2𝛼+𝛽 Taking (β – α ) common from R1 = (α + β + 𝛾)(β – α) 1a2β+𝛾0𝛽+𝛼−11y2𝛼+𝛽 Applying R3 → R3 − R1 = (α + β + 𝛾)(β – α) 1a2β+𝛾0𝛽+𝛼−1𝟏−𝟏y2−𝛼2𝛼+𝛽−𝛽−𝛾 = (α + β + 𝛾)(β – α) 1a2β+𝛾0(𝛽+𝛼)−1𝟎(𝛾−𝛼)(𝛾+𝛼)−(𝛾−𝛼) Taking (𝛾 – α) common from R3 = (α + β + 𝛾)(β – α) (𝛾 – α) 1a2β+𝛾0𝛽+𝛼−10𝛾+𝛼−1 Expanding determinant along C1 = (α + β + 𝛾)(β – α) (𝛾 – α) 1 𝛽+𝛼−1𝛾+𝛼−1−0 𝛼2𝛽+𝛾𝛾+𝛼−1+0 𝛼2𝛽+𝛾𝛽+𝛼−1 = (α + β + 𝛾)(β – α) (𝛾 – α) 1 𝛽+𝛼−1𝛾+𝛼−1−0+0 = (α + β + 𝛾)(β – α) (𝛾 – α) ( – (β + α ) + 𝛾 + α ) – 0 + 0) = (α + β + 𝛾)(β – α) (𝛾 – α) ( – β – α + 𝛾 + α ) = (α + β + 𝛾)(β – α) (𝛾 – α) ( – β + 𝛾 ) = (α + β + 𝛾)(β – α) (𝛾 – α) (𝛾 – β ) = (α + β + 𝛾)(β – α) (𝛾 – α) (β – 𝛾 ) = (α + β + 𝛾)(–(α –β)) (–(α – 𝛾)) (β – 𝛾 ) = (α + β + 𝛾) (α – β) (α – 𝛾) (β – 𝛾) = 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 You are here

Misc 12 Important

Misc. 13

Misc 14

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.