Question 3 - Miscellaneous - Chapter 4 Class 12 Determinants
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 3 Solve the equations 𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 = 0, a ≠ 0 Solving ∆ = 𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Applying R1 → R1 + R2 + R3 = 𝑥+𝑎+𝑥+𝑥𝑥+𝑥+𝑎+𝑥𝑥+𝑥+𝑥+𝑎𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 = 𝟑𝒙+𝒂𝟑𝒙+𝒂𝟑𝒙+𝒂𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Taking (3x + a) common from R1 = (3x + a) 111𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Applying C2 → C2 – C1 = (3x + a) 1𝟏−𝟏1𝑥𝑥+𝑎−𝑥𝑥𝑥𝑥−𝑥𝑥+𝑎 = (3x+ a) 1𝟎1𝑥𝑎𝑥𝑥0𝑥+𝑎 Applying C3 →C3 – C1 = (3x + a) 10𝟏−𝟏𝑥𝑎𝑥−𝑥𝑥0𝑥+𝑎−𝑥 = (3x + a) 10𝟎𝑥𝑎0𝑥0𝑎 = (3x+ a) 1 𝑎00𝑎−0 𝑥0𝑥0+0 𝑥𝑎𝑥0 = (3x+ a) 1 𝑎00𝑎−0+0 = (3x+ a) (1(a2 – 0) – 0 + 0) = (3x+ a) (a2) ∴ ∆ = (3x+ a) (a2) Given ∆ = 0 ∴ (3x + a) a2 = 0
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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