Ex 3.2
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5
Ex 3.2, 6
Ex 3.2, 7 (i)
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 18 You are here
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Last updated at April 16, 2024 by Teachoo
Ex 3.2, 18 If A =[■8(0&−tan 𝛼/2 " " @tan 𝛼/2 " " &0)] and I is the identity matrix of order 2, Show that I + A = ( I – A)[■8(cos𝛼&−sin𝛼@sin𝛼&cos𝛼 )] Given I the identity matrix of order 2 i.e. I = [■8(1&0@0&1)] Solving L.H.S. I + A = [■8(1&0@0&1)] + [■8(0&−tan 𝛼/2 " " @tan 𝛼/2 " " &0)] = [■8(1+0&0−tan 𝛼/2 " " @0+tan 𝛼/2 " " &1+0)] = [■8(𝟏&−𝐭𝐚𝐧 𝜶/𝟐 " " @𝐭𝐚𝐧 𝜶/𝟐 " " &𝟏)] Solving R.H.S (I – A) [■8(cos𝛼&−sin𝛼@sin𝛼&cos𝛼 )] = (" " [■8(1&0@0&1)]−[■8(0&−tan 𝛼/2 " " @tan 𝛼/2 " " &0)]) [■8(cos𝛼&−sin𝛼@sin𝛼&cos𝛼 )] = [■8(1−0&0+tan 𝛼/2 " " @0−tan 𝛼/2 " " &1)][■8(cos𝛼&−sin𝛼@sin𝛼&cos𝛼 )] = [■8(1&tan α/2 " " @− tan α/2 " " &1)][■8(cos𝛼&−sin𝛼@sin𝛼&cos𝛼 )] = [■8(1(𝑐𝑜𝑠𝛼 )+𝑡𝑎𝑛 𝛼/2 (𝑠𝑖𝑛𝛼)&1(〖−𝑠𝑖𝑛〗𝛼 )+𝑡𝑎𝑛 𝛼/2 (𝑠𝑖𝑛𝛼)@−𝑡𝑎𝑛 𝛼/2 (𝑐𝑜𝑠𝛼 )+1(𝑠𝑖𝑛𝛼) &−𝑡𝑎𝑛 𝛼/2 (〖−𝑠𝑖𝑛〗𝛼 )+1(𝑠𝑖𝑛𝛼))] = [■8(1((1 − 𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2))"+ " 𝑡𝑎𝑛 𝛼/2 ((2 tan〖 𝛼/2〗)/(1 + 𝑡𝑎𝑛2 𝛼/2))" " &1((−2 tan〖 𝛼/2〗)/(1 + 𝑡𝑎𝑛2 ( 𝛼)/2))"+(" 𝑡𝑎𝑛 𝛼/2) ((1 − 𝑡𝑎𝑛2 ( 𝛼)/2)/(1 + 𝑡𝑎𝑛2 ( 𝛼)/2))" " @−𝑡𝑎𝑛 𝛼/2 ((1 − 𝑡𝑎𝑛2 ( 𝛼)/2)/(1 + 𝑡𝑎𝑛2 ( 𝛼)/2))" +1" ((2 tan〖 𝛼/2〗)/(1 + 𝑡𝑎𝑛2 𝛼/2))&−𝑡𝑎𝑛 𝛼/2 ((−2 〖tan 〗〖𝛼/2〗)/(1 + 𝑡𝑎𝑛2 𝛼/2))" +1" ((1 − 𝑡𝑎𝑛2 ( 𝛼)/2)/(1 + 𝑡𝑎𝑛2 𝛼/2)) )] We know that cos 2θ = (1 − 𝑡𝑎𝑛2𝜃)/(1 + 𝑡𝑎𝑛2𝜃) & sin 2θ = (2 tan𝜃)/(1 + 𝑡𝑎𝑛2𝜃) Replacing θ with 𝜃/2 So, cos θ = (1 − 𝑡𝑎𝑛2 𝜃/2)/(1 + 𝑡𝑎𝑛2 𝜃/2) & sin θ = (2 𝑡𝑎𝑛〖 𝜃/2〗)/(1 + 𝑡𝑎𝑛2 𝜃/2) = [■8((1 − 𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2) " +" (2𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2) " " &(−2 tan〖 𝛼/2〗 " +" tan〖 𝛼/2〗 − 𝑡𝑎𝑛3 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2) " " @(−〖tan 〗〖𝛼/2〗 (1 − 𝑡𝑎𝑛2 𝛼/2))/(1 + 𝑡𝑎𝑛2 𝛼/2) " +" (2 tan〖 𝛼/2〗)/(1 + 𝑡𝑎𝑛2 𝛼/2)&(2𝑡𝑎𝑛2 𝛼/2)/(1+𝑡𝑎𝑛2 𝛼/2)+ " " (1 − 𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2))] = [■8((1 − 𝑡𝑎𝑛2 𝛼/2 + 2𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2) " " &(−2 〖tan 〗〖 𝛼/2〗 " + " tan〖 𝛼/2〗 − 𝑡𝑎𝑛3 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2) " " @(−〖tan 〗〖𝛼/2〗 + 𝑡𝑎𝑛3 𝛼/2 +2 tan〖 𝛼/2〗)/(1 + 𝑡𝑎𝑛2 𝛼/2) " " &" " (2𝑡𝑎𝑛2 𝛼/2 + 1 − 𝑡𝑎𝑛2 𝛼/2)/(1 + 𝑡𝑎𝑛2 𝛼/2))] = [■8((1 + 𝑡𝑎𝑛2 𝛼/2 )/(1 + 𝑡𝑎𝑛2 𝛼/2) " " &(−〖tan 〗〖𝛼/2〗 " " (1 + 𝑡𝑎𝑛2 𝛼/2) )/(1 + 𝑡𝑎𝑛2 𝛼/2) " " @(〖tan 〗〖𝛼/2〗 (1 + 𝑡𝑎𝑛2 𝛼/2))/(1 + 𝑡𝑎𝑛2 𝛼/2) " " &" " (1 + 𝑡𝑎𝑛2 𝛼/2 )/(1 + 𝑡𝑎𝑛2 ( 𝛼)/2))] = [■8(𝟏&−𝐭𝐚𝐧 𝜶/𝟐 " " @𝐭𝐚𝐧 𝜶/𝟐 " " &𝟏)] = R.H.S. Hence proved