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Misc 11 - Using properties of determinants - Determinants

Misc 11 - Chapter 4 Class 12 Determinants - Part 2
Misc 11 - Chapter 4 Class 12 Determinants - Part 3
Misc 11 - Chapter 4 Class 12 Determinants - Part 4

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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.