# Misc 11

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

Ex 4.1, 7
Important

Example 14 Important

Example 15 Important

Example 16 Important

Ex 4.2, 7 Important

Ex 4.2, 8 Important

Ex 4.2, 11 Important

Ex 4.2, 12 Important

Ex 4.2, 13 Important

Ex 4.2, 14 Important

Ex 4.2, 15 Important

Example 18 Important

Ex 4.3, 2 Important

Ex 4.3, 3 Important

Example 24 Important

Example 26 Important

Ex 4.5, 10 Important

Ex 4.5, 15 Important

Ex 4.5, 18 Important

Ex 4.6, 13 Important

Ex 4.6, 15 Important

Ex 4.6, 16 Important

Example 32 Important

Example 34 Important

Misc. 2 Important

Misc 11 Important You are here

Misc. 15 Important

Misc. 16 Important

Misc. 19 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.