1. Class 12
2. Important Question for exams Class 12
3. Chapter 4 Class 12 Determinants

Transcript

Misc 11 Using properties of determinants, prove that: 𝛼﷮a2﷮β+𝛾﷮β﷮β2﷮𝛾+𝛼﷮𝛾﷮𝛾2﷮𝛼+β﷯﷯ = (β – 𝛾) (𝛾 – 𝛼) (𝛼 – β) (a + β + 𝛾) Taking L.H.S 𝛼﷮𝛼2﷮β+y﷮β﷮β2﷮y+𝛼﷮y﷮y2﷮𝛼+β﷯﷯ Applying C1→ C1 + C3 = 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) 1﷮𝛼2﷮β+𝛾﷮1﷮β2﷮𝛾+𝛼﷮1﷮𝛾2﷮𝛼+β﷯﷯ Applying R2→ R2 – R1 = (α + β + 𝛾) 1﷮a2﷮β+𝛾﷮𝟏−𝟏﷮β2−a2﷮𝛾+𝛼−𝛽−𝛾﷮1﷮y2﷮𝛼+𝛽﷯﷯ = (α + β + 𝛾) 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) 1﷮a2﷮β+𝛾﷮0﷮𝛽+𝛼﷮−1﷮1﷮y2﷮𝛼+𝛽﷯﷯ Applying R3 → R3 − R1 = (α + β + 𝛾)(β – α) 1﷮a2﷮β+𝛾﷮0﷮𝛽+𝛼﷮−1﷮𝟏−𝟏﷮y2−𝛼2﷮𝛼+𝛽−𝛽−𝛾﷯﷯ = (α + β + 𝛾)(β – α) 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) 1﷮a2﷮β+𝛾﷮0﷮𝛽+𝛼﷮−1﷮0﷮𝛾+𝛼﷮−1﷯﷯ Expanding determinant along C1 = (α + β + 𝛾)(β – α) (𝛾 – α) 1 𝛽+𝛼﷮−1﷮𝛾+𝛼﷮−1﷯﷯−0 𝛼2﷮𝛽+𝛾﷮𝛾+𝛼﷮−1﷯﷯+0 𝛼2﷮𝛽+𝛾﷮𝛽+𝛼﷮−1﷯﷯﷯ = (α + β + 𝛾)(β – α) (𝛾 – α) 1 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

Chapter 4 Class 12 Determinants

Class 12
Important Question for exams Class 12