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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

Transcript

Misc 1 Prove that determinant |β– 8(π‘₯&π‘ π‘–π‘›β‘πœƒ&π‘π‘œπ‘ β‘πœƒ@βˆ’π‘ π‘–π‘›β‘πœƒ&βˆ’π‘₯&1@π‘π‘œπ‘ β‘πœƒ&1&π‘₯)| is independent of ΞΈ. Let βˆ† = |β– 8(π‘₯&π‘ π‘–π‘›β‘πœƒ&π‘π‘œπ‘ β‘πœƒ@βˆ’π‘ π‘–π‘›β‘πœƒ&βˆ’π‘₯&1@π‘π‘œπ‘ β‘πœƒ&1&π‘₯)| βˆ† = x |β– 8(βˆ’π‘₯&1@1&π‘₯)| – sin ΞΈ |β– 8(βˆ’sin⁑θ&1@cos⁑θ&π‘₯)| + cos ΞΈ |β– 8(βˆ’sin⁑θ&βˆ’π‘₯@cos⁑θ&1)| = x ( –x2 – 1) – sin ΞΈ ( –xsin ΞΈ – cos ΞΈ) + cos ΞΈ (–sin ΞΈ + x cos ΞΈ) = –x3 – x + x sin⁑〖2 ΞΈγ€— + 𝐬𝐒𝐧⁑𝛉 cos ΞΈ – sin ΞΈ cos ΞΈ + x cos2 ΞΈ = –x3 – x + x sin2 ΞΈ + x cos2 ΞΈ = –x3 – x + x (sin2 ΞΈ + cos2 ΞΈ) = –x3 – x + x (1) = –x3 (As sin2 ΞΈ + cos2 ΞΈ = 1) Hence βˆ† = –x3 Which has no ΞΈ term Thus, the determinant is independent of ΞΈ Hence Proved

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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.