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  1. Chapter 3 Class 12 Matrices
  2. Serial order wise

Transcript

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(π‘π‘œπ‘ β‘π›Ό )+π‘‘π‘Žπ‘› 𝛼/2 (𝑠𝑖𝑛⁑𝛼)&1(γ€–βˆ’π‘ π‘–π‘›γ€—β‘π›Ό )+π‘‘π‘Žπ‘› 𝛼/2 (𝑠𝑖𝑛⁑𝛼)@βˆ’π‘‘π‘Žπ‘› 𝛼/2 (π‘π‘œπ‘ β‘π›Ό )+1(𝑠𝑖𝑛⁑𝛼) &βˆ’π‘‘π‘Žπ‘› 𝛼/2 (γ€–βˆ’π‘ π‘–π‘›γ€—β‘π›Ό )+1(𝑠𝑖𝑛⁑𝛼))] = [β– 8(1((1 βˆ’ π‘‘π‘Žπ‘›2 𝛼/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))"+ " π‘‘π‘Žπ‘› 𝛼/2 ((2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" " &1((βˆ’2 tan⁑〖 𝛼/2γ€—)/(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" ((2 tan⁑〖 𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))&βˆ’π‘‘π‘Žπ‘› 𝛼/2 ((βˆ’2 γ€–tan 〗⁑〖𝛼/2γ€—)/(1 + π‘‘π‘Žπ‘›2 𝛼/2))" +1" ((1 βˆ’ π‘‘π‘Žπ‘›2 ( 𝛼)/2)/(1 + π‘‘π‘Žπ‘›2 𝛼/2)) )] We know that cos 2ΞΈ = (1 βˆ’ π‘‘π‘Žπ‘›2πœƒ)/(1 + π‘‘π‘Žπ‘›2πœƒ) & sin 2ΞΈ = (2 tanβ‘πœƒ)/(1 + π‘‘π‘Žπ‘›2πœƒ) Replacing ΞΈ with πœƒ/2 So, cos ΞΈ = (1 βˆ’ π‘‘π‘Žπ‘›2 πœƒ/2)/(1 + π‘‘π‘Žπ‘›2 πœƒ/2) & sin ΞΈ = (2 π‘‘π‘Žπ‘›β‘γ€– πœƒ/2γ€—)/(1 + π‘‘π‘Žπ‘›2 πœƒ/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. 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.