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Last updated at Jan. 23, 2020 by Teachoo

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Misc 11 Using properties of determinants, prove that: |■8(𝛼&∝^2&β+𝛾@β&β2&𝛾+𝛼@𝛾&𝛾2&𝛼+β)| = (β – 𝛾) (𝛾 – 𝛼) (𝛼 – β) (a + β + 𝛾) Solving L.H.S |■8(𝛼&∝^2&β+y@β&β2&y+𝛼@y&y2&𝛼+β)| Applying C1→ C1 + C3 = |■8(𝜶+𝜷+𝜸&𝛼2&β+𝛾@𝛃+𝜸+𝜶&β2&𝛾+𝛼@𝜸+𝜶+𝜷&𝛾2&𝛼+β)| Taking (α + β + 𝜸) common from C1 = (α + β + 𝜸) |■8(1&𝛼2&β+𝛾@1&β2&𝛾+𝛼@1&𝛾2&𝛼+β)| Applying R2→ R2 – R1 = (α + β + 𝛾) |■8(1&a2&β+𝛾@𝟏−𝟏&β2−a2&𝛾+𝛼−𝛽−𝛾@1&y2&𝛼+𝛽)| = (α + β + 𝛾) |■8(1&a2&β+𝛾@𝟎&(β−a)(𝛽+𝛼)&−(𝛽−𝛼)@1&y2&𝛼+𝛽)| Taking (β – α ) common from R1 = (α + β + 𝛾)(β – α) |■8(1&a2&β+𝛾@0&𝛽+𝛼&−1@1&y2&𝛼+𝛽)| Applying R3 → R3 − R1 = (α + β + 𝛾)(β – α) |■8(1&a2&β+𝛾@0&𝛽+𝛼&−1@𝟏−𝟏&y2−𝛼2&𝛼+𝛽−𝛽−𝛾)| = (α + β + 𝛾)(β – α) |■8(1&a2&β+𝛾@0&(𝛽+𝛼)&−1@𝟎&(𝛾−𝛼)(𝛾+𝛼)&−(𝛾−𝛼))| Taking (𝛾 – α) common from R3 = (α + β + 𝛾)(β – α) (𝛾 – α) |■8(1&a2&β+𝛾@0&𝛽+𝛼&−1@0&𝛾+𝛼&−1)| Expanding determinant along C1 = (α + β + 𝛾)(β – α) (𝛾 – α)(1|■8(𝛽+𝛼&−1@𝛾+𝛼&−1)|−0|■8(𝛼2&𝛽+𝛾@𝛾+𝛼&−1)|+0|■8(𝛼2&𝛽+𝛾@𝛽+𝛼&−1)|) = (α + β + 𝛾)(β – α) (𝛾 – α)(1|■8(𝛽+𝛼&−1@𝛾+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.