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Ex 3.2, 18 - Show that I + A = (I - A) [cos a -sin a - Solving Equation

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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Ex 3.2, 18 If A =[β– 8(0&βˆ’tan 𝛼/2 " " @tan 𝛼/2 " " &0)] and I is the identity matrix of order 2, Show that I + A = ( I – A)[β– 8(cos⁑𝛼&βˆ’sin⁑𝛼@sin⁑𝛼&cos⁑𝛼 )] Given I the identity matrix of order 2 i.e. I = [β– 8(1&0@0&1)] Taking L.H.S. I + A = [β– 8(1&0@0&1)] + [β– 8(0&βˆ’tan 𝛼/2 " " @tan 𝛼/2 " " &0)] = [β– 8(1+0&0βˆ’tan 𝛼/2 " " @0+tan 𝛼/2 " " &1+0)] = [β– 8(1&βˆ’tan 𝛼/2 " " @tan 𝛼/2 " " &1)] Taking R.H.S (I – A) [β– 8(cos⁑𝛼&βˆ’sin⁑𝛼@sin⁑𝛼&cos⁑𝛼 )] = (" " [β– 8(1&0@0&1)]βˆ’[β– 8(0&βˆ’tan 𝛼/2 " " @tan 𝛼/2 " " &0)]) [β– 8(cos⁑𝛼&βˆ’sin⁑𝛼@sin⁑𝛼&cos⁑𝛼 )] = [β– 8(1βˆ’0&0+tan 𝛼/2 " " @0βˆ’tan 𝛼/2 " " &1)][β– 8(cos⁑𝛼&βˆ’sin⁑𝛼@sin⁑𝛼&cos⁑𝛼 )] = [β– 8(1&tan Ξ±/2 " " @βˆ’ tan Ξ±/2 " " &1)][β– 8(cos⁑𝛼&βˆ’sin⁑𝛼@sin⁑𝛼&cos⁑𝛼 )] = [β– 8(1&tan Ξ±/2 " " @βˆ’ tan Ξ±/2 " " &1)] [β– 8((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &(βˆ’2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " @(2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &(1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " )] = [β– 8(1((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))"+ " π‘‘π‘Žπ‘› 𝛼/2 ((2π‘‘π‘Žπ‘› 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" " &1((βˆ’2π‘‘π‘Žπ‘› 𝛼/2)/(1+π‘‘π‘Žπ‘›2 𝛼/2))"+(" π‘‘π‘Žπ‘› 𝛼/2) ((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" " @βˆ’π‘‘π‘Žπ‘› 𝛼/2 ((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" +1" ((2π‘‘π‘Žπ‘› 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))&βˆ’π‘‘π‘Žπ‘› 𝛼/2 ((βˆ’2π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" +1" ((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2)) )] = [β– 8((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " +" (2π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &(βˆ’2 tan⁑〖 𝛼/2γ€— " +" tan⁑〖 𝛼/2γ€— βˆ’ π‘‘π‘Žπ‘›3 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " @(βˆ’γ€–tan 〗⁑〖𝛼/2γ€— (1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2))/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " +" (2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2)&(2π‘‘π‘Žπ‘›2 𝛼/2)/(1+π‘‘π‘Žπ‘›2 𝛼/2)+ " " (1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))] = [β– 8((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2 + 2π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &(βˆ’2 γ€–tan 〗⁑〖𝛼/2γ€— " + " tan⁑〖 𝛼/2γ€— βˆ’ π‘‘π‘Žπ‘›3 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " @(βˆ’γ€–tan 〗⁑〖𝛼/2γ€— + π‘‘π‘Žπ‘›3 𝛼/2+2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &" " (2π‘‘π‘Žπ‘›2 𝛼/2 + 1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))] = [β– 8((1 + π‘‘π‘Žπ‘›2 𝛼/2 )/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &(βˆ’γ€–tan 〗⁑〖𝛼/2γ€— " " (1 + π‘‘π‘Žπ‘›2 𝛼/2) )/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " @(γ€–tan 〗⁑〖𝛼/2γ€— (1 + π‘‘π‘Žπ‘›2 𝛼/2))/(1 + π‘‘π‘Žπ‘›2 𝛼/2) " " &" " (1 + π‘‘π‘Žπ‘›2 𝛼/2 )/(1 + π‘‘π‘Žπ‘›2 ( 𝛼)/2))] = [β– 8(1&βˆ’tan 𝛼/2 " " @tan 𝛼/2 " " &1)] = R.H.S.

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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