# Ex 4.5, 11 - Chapter 4 Class 12 Determinants

Last updated at Jan. 22, 2020 by Teachoo

Last updated at Jan. 22, 2020 by Teachoo

Transcript

Ex 4.5, 11 Find the inverse of each of the matrices [■8(1&0&0@0&cos𝛼&sin𝛼@0&sin𝛼&−cos𝛼 )] Let A =[■8(1&0&0@0&cos𝛼&sin𝛼@0&sin𝛼&−cos𝛼 )] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Calculating |A| |A| = |■8(1&0&0@0&cos𝛼&sin𝛼@0&sin𝛼&−cos𝛼 )| = 1 |■8(cos𝛼&sin𝛼@sin𝛼&−cos𝛼 )|– 0 |■8(0&sin𝛼@0&〖− cos〗𝛼 )|+ 0|■8(0&cos𝛼@0&𝑠𝑖𝑛 𝛼)| = 1(– cos2α – sin2α ) – 0 + 0 = –( cos2α + sin2α ) = –1 Since |𝐴|≠ 0 Thus A-1 exists Calculating adj A adj (A) = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(1&0&0@0&cos𝛼&sin𝛼@0&sin𝛼&−cos𝛼 )] M11 = |■8(cos"α" &sin"α" @sin"α" &−cos"α" )| = –cos2α – sin2α = –(cos^2α 〖+ 𝑠𝑖𝑛〗^2α ) = –1 M12 = |■8(0&sin 𝛼@0&−cos 𝛼)| = 0 – 0 = 0 M13 = |■8(0&cos𝛼@0&sin 𝛼)| = 0 – 0 = 0 M21 = |■8(0&0@sin 𝛼&−cos𝛼 )| = 0 – 0 = 0 M22 = |■8(1&0@0&−cos 𝛼)| = –cos α – 0 = –cos α M23 = |■8(1&0@0&sin𝛼 )| = sin α = 0 = sin α M31 = |■8(0&0@cos 𝛼&sin 𝛼)| = 0 – 0 = –0 M32 = |■8(1&0@0&sin 𝛼)| = sin α – 0 = sin α M33 = |■8(1&0@0&cos 𝛼)| = cos α + 0 = cos α Now, A11 = 〖(−1)〗^(1+1) M11 = 〖(−1)〗^2 (–1)2 = –1 A12 = 〖(−1)〗^(1+2) M12 = 〖(−1)〗^3 0 = 0 A13 = 〖(−1)〗^(1+3) M13 = 〖(−1)〗^4 0 = 0 A21 = 〖(−1)〗^(2+1)M21 = (–1)3 0 = 0 A22 = 〖(−1)〗^(2+2) M22 = 〖(−1)〗^4(– cos α) = –cos α A23 = 〖"(– 1)" 〗^(2+3) M23 = 〖"(–1)" 〗^5 sin α = –sin α A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 0 = 0 A32 = 〖(−1)〗^(3+2)sin α = (–1)5 sin α = –sin α A33 = 〖(−1)〗^(3+3)M33 = (–1)6 cos α = cos α So, adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(−1&0&0@0&−cos𝛼&−sin𝛼@0&−sin𝛼&cos𝛼 )] Calculating inverse Now, A– 1 = 1/(|A|) ( adj (A)) = 1/(−1) [■8(−1&0&0@0&−cos𝛼&−sin𝛼@0&−sin𝛼&cos𝛼 )] = – [■8(−1&0&0@0&−cos𝛼&−sin𝛼@0&−sin𝛼&cos𝛼 )] = [■8(𝟏&𝟎&𝟎@𝟎&𝒄𝒐𝒔𝜶&𝒔𝒊𝒏𝜶@𝟎&𝒔𝒊𝒏𝜶&〖−𝒄𝒐𝒔〗𝜶 )]

Chapter 4 Class 12 Determinants

Serial order wise

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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.