Last updated at Dec. 16, 2024 by Teachoo
Ex 10.5, 1 (Supplementary NCERT) Find [π β" " π β" " π β ] if π β = π Μ β 2π Μ + 3π Μ , π β = 2π Μ β 3π Μ + π Μ , π β = 3π Μ + π Μ β 2π ΜGiven, π β = π Μ β 2π Μ + 3π Μ , π β = 2π Μ β 3π Μ + π Μ , π β = 3π Μ + π Μ β 2π Μ Now, π β.(π β Γ π β) = [π β" " π β" " π β ] = |β 8(π&βπ&π@π&βπ&π@π&π&βπ)| = 1[(β3Γβ2)β(1Γ1) ] β (β2)[(2Γβ2)β(3Γ1) ] + 3[(2Γ1)β(3Γβ3)] = 1 [6β1]+2(β4β3)+3[2+9] = 1(5) + 2(β7) + 3(11) = 5 β 14 + 33 = 38 β 14 = 24 β΄ [π β" " π β" " π β ] = 24
Ex 10.5 (Supplementary NCERT)
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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