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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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)] Solving 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’π­πšπ§ 𝜢/𝟐 " " @𝐭𝐚𝐧 𝜢/𝟐 " " &𝟏)] Solving 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’π­πšπ§ 𝜢/𝟐 " " @𝐭𝐚𝐧 𝜢/𝟐 " " &𝟏)] = R.H.S. 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.