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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 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo