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Ex 4.5, 11 - Find inverse [1 0 0 0 cos sin 0 sin -cos] - Ex 4.5

Ex 4.5, 11 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 11 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.5, 11 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.5, 11 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.5, 11 - Chapter 4 Class 12 Determinants - Part 6

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Transcript

Ex 4.5, 11 Find the inverse of each of the matrices [■8(1&0&0@0&cos⁡𝛼&sin⁡𝛼@0&sin⁡𝛼&−cos⁡𝛼 )] Let A =[■8(1&0&0@0&cos⁡𝛼&sin⁡𝛼@0&sin⁡𝛼&−cos⁡𝛼 )] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Calculating |A| |A| = |■8(1&0&0@0&cos⁡𝛼&sin⁡𝛼@0&sin⁡𝛼&−cos⁡𝛼 )| = 1 |■8(cos⁡𝛼&sin⁡𝛼@sin⁡𝛼&−cos⁡𝛼 )|– 0 |■8(0&sin⁡𝛼@0&〖− cos〗⁡𝛼 )|+ 0|■8(0&cos⁡𝛼@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&cos⁡𝛼&sin⁡𝛼@0&sin⁡𝛼&−cos⁡𝛼 )] 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&cos⁡𝛼@0&sin 𝛼)| = 0 – 0 = 0 M21 = |■8(0&0@sin 𝛼&−cos⁡𝛼 )| = 0 – 0 = 0 M22 = |■8(1&0@0&−cos 𝛼)| = –cos α – 0 = –cos α M23 = |■8(1&0@0&sin⁡𝛼 )| = 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&−cos⁡𝛼&−sin⁡𝛼@0&−sin⁡𝛼&cos⁡𝛼 )] Calculating inverse Now, A– 1 = 1/(|A|) ( adj (A)) = 1/(−1) [■8(−1&0&0@0&−cos⁡𝛼&−sin⁡𝛼@0&−sin⁡𝛼&cos⁡𝛼 )] = – [■8(−1&0&0@0&−cos⁡𝛼&−sin⁡𝛼@0&−sin⁡𝛼&cos⁡𝛼 )] = [■8(𝟏&𝟎&𝟎@𝟎&𝒄𝒐𝒔⁡𝜶&𝒔𝒊𝒏⁡𝜶@𝟎&𝒔𝒊𝒏⁡𝜶&〖−𝒄𝒐𝒔〗⁡𝜶 )]

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.