# 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

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 9 years. He provides courses for Maths and Science at Teachoo.