Check sibling questions

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

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

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(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] Given I the identity matrix of order 2 i.e. I = [■8(1&[email protected]&1)] Taking L.H.S. I + A = [■8(1&[email protected]&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)] Taking R.H.S (I – A) [■8(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] = (" " [■8(1&[email protected]&1)]−[■8(0&−tan 𝛼/2 " " @tan 𝛼/2 " " &0)]) [■8(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] = [■8(1−0&0+tan 𝛼/2 " " @0−tan 𝛼/2 " " &1)][■8(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] = [■8(1&tan α/2 " " @− tan α/2 " " &1)][■8(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] = [■8(1(𝑐𝑜𝑠⁡𝛼 )+𝑡𝑎𝑛 𝛼/2 (𝑠𝑖𝑛⁡𝛼)&1(〖−𝑠𝑖𝑛〗⁡𝛼 )+𝑡𝑎𝑛 𝛼/2 (𝑠𝑖𝑛⁡𝛼)@−𝑡𝑎𝑛 𝛼/2 (𝑐𝑜𝑠⁡𝛼 )+1(𝑠𝑖𝑛⁡𝛼) &−𝑡𝑎𝑛 𝛼/2 (〖−𝑠𝑖𝑛〗⁡𝛼 )+1(𝑠𝑖𝑛⁡𝛼))] = [■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 tan⁡𝜃)/(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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.