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

Ex 3.2, 18 - Chapter 3 Class 12 Matrices - Part 2
Ex 3.2, 18 - Chapter 3 Class 12 Matrices - Part 3 Ex 3.2, 18 - Chapter 3 Class 12 Matrices - Part 4 Ex 3.2, 18 - Chapter 3 Class 12 Matrices - Part 5

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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&[email protected]&1)] Taking L.H.S. I + A = [β– 8(1&[email protected]&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&[email protected]&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 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.