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Ex 4.5, 7 - Find inverse of matrix [ 1 2 3 0 2 4 0 0 5] - Teachoo

Ex 4.5, 7 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 7 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.5, 7 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.5, 7 - Chapter 4 Class 12 Determinants - Part 5

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Ex 4.5, 7 Find the inverse of each of the matrices (if it exists).[■8(1&2&3@0&2&4@0&0&5)] Let A = [■8(1&2&3@0&2&4@0&0&5)] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Calculating |A| |A| = |■8(1&2&3@0&2&4@0&0&5)| = 1 |■8(2&4@0&5)|– 2 |■8(0&4@0&5)| – 3|■8(0&2@0&0)| = 1(10 – 0) –2 (0 – 0) –3 (0 – 0) = 10 – 0 – 0 = 10 Since |A| ≠ 0 ∴ A–1 exists Calculating adj A adj A = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] A = [■8(1&2&3@0&2&4@0&0&5)] M11 = |■8(2&4@0&5)| = 2(5) – 0(4) = 10 M12 = |■8(0&4@0&5)| = 0(5) – 0(4) = 0 M13 = |■8(0&2@0&0)| = 0 – 0 = 0 M21 = |■8(2&3@0&5)| = 10 – 0 = 10 M22 = |■8(1&3@0&5)| = 5 – 0 = 5 M23 = |■8(1&2@0&0)| = 0 – 0 = 0 M31 = |■8(2&3@2&4)| = 8 – 6 = 2 M32 = |■8(1&3@0&4)| = 4 – 0 = 4 M33 = |■8(1&2@0&2)| = 2 – 0 = 2 Now, A11 = (–1)1-1 M11 = (–1)2 (10) = 1(10) = 10 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 (10) = –1 (10) = –10 A22 = (–1)2+2 M22 = (–1)4 5 = 5 A23 = (–1)2+3 M23 = (–1)4 0 = 0 A31 = (–1)3+1 M31 = (–1)4 2 = 2 A32 = (–1)3+2 M32 = (–1)5 4= –1 (4) = –4 A33 = (–1)3+3 M33 = (–1)6 2 = 2 Thus adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)]= [■8(10&−10&2@0&5&−4@0&0&2)] Calculating inverse Now , A– 1 = 1/(|A|) ( adj (A)) = 𝟏/𝟏𝟎 [■8(𝟏𝟎&−𝟏𝟎&𝟐@𝟎&𝟓&−𝟒@𝟎&𝟎&𝟐)]

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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.