Get live Maths 1-on-1 Classs - Class 6 to 12

Miscellaneous

Misc 1

Misc. 2 Important Deleted for CBSE Board 2023 Exams

Misc 3

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 Important Deleted for CBSE Board 2023 Exams You are here

Misc 12 Important Deleted for CBSE Board 2023 Exams

Misc. 13 Deleted for CBSE Board 2023 Exams

Misc 14 Deleted for CBSE Board 2023 Exams

Misc. 15 Important Deleted for CBSE Board 2023 Exams

Misc. 16 Important

Misc 17 (MCQ) Important Deleted for CBSE Board 2023 Exams

Misc 18 (MCQ)

Misc 19 (MCQ) Important

Matrices and Determinants - Formula Sheet and Summary Important

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at March 16, 2023 by Teachoo

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[email protected]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&−[email protected]&y2&𝛼+𝛽)| Applying R3 → R3 − R1 = (α + β + 𝛾)(β – α) |■8(1&a2&β+𝛾@0&𝛽+𝛼&−[email protected]𝟏−𝟏&y2−𝛼2&𝛼+𝛽−𝛽−𝛾)| = (α + β + 𝛾)(β – α) |■8(1&a2&β+𝛾@0&(𝛽+𝛼)&−[email protected]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&−[email protected]&𝛾+𝛼&−1)| Expanding determinant along C1 = (α + β + 𝛾)(β – α) (𝛾 – α)(1|■8(𝛽+𝛼&−[email protected]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&−[email protected]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