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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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Ex 4.5, 11 Find the inverse of each of the matrices [■8(1&0&[email protected]&cos⁡𝛼&sin⁡𝛼@0&sin⁡𝛼&−cos⁡𝛼 )] Let A =[■8(1&0&[email protected]&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&[email protected]&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&[email protected]&A22&[email protected]&A23&A33)] A = [■8(1&0&[email protected]&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&[email protected] 𝛼&−cos⁡𝛼 )| = 0 – 0 = 0 M22 = |■8(1&[email protected]&−cos 𝛼)| = –cos α – 0 = –cos α M23 = |■8(1&[email protected]&sin⁡𝛼 )| = sin α = 0 = sin α M31 = |■8(0&[email protected] 𝛼&sin 𝛼)| = 0 – 0 = –0 M32 = |■8(1&[email protected]&sin 𝛼)| = sin α – 0 = sin α M33 = |■8(1&[email protected]&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&[email protected]&A22&[email protected]&A23&A33)] = [■8(−1&0&[email protected]&−cos⁡𝛼&−sin⁡𝛼@0&−sin⁡𝛼&cos⁡𝛼 )] Calculating inverse Now, A– 1 = 1/(|A|) ( adj (A)) = 1/(−1) [■8(−1&0&[email protected]&−cos⁡𝛼&−sin⁡𝛼@0&−sin⁡𝛼&cos⁡𝛼 )] = – [■8(−1&0&[email protected]&−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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.