Slide25.JPG

Slide26.JPG
Slide27.JPG
Slide28.JPG

Go Ad-free

Transcript

Ex 4.4, 8 Find the inverse of each of the matrices (if it exists). [■8(1&0&0@3&3&0@5&2&−1)] Let A = [■8(1&0&0@3&3&0@5&2&−1)] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = |■8(1&0&0@3&3&0@5&2&−1)| = 1 |■8(3&0@2&−1)| – 0 |■8(3&0@5&1)| + 0|■8(3&3@5&2)| = 1(–3 – 0) + 0 + 0 = −3 Since |A| ≠ 0 , A–1 exists Step 2: Calculate adj A adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] A = [■8(1&0&0@3&3&0@5&2&−1)] M11 = |■8(3&0@2&−1)| = 3(−1) – 2(0) = −3 M12 = |■8(3&0@5&−1)| = 3(–1) – 0(5) = – 3 M13 = |■8(3&3@5&2)| = 3(2) – 5(3) = – 9 M21 = |■8(0&0@2&–1)| = 0(–1) – 2(0) = 0 M22 = |■8(1&0@5&−1)| = 1(– 1) – 0 = – 1 M23 = |■8(1&0@5&2)| = 2(1) – 5(0) = 2 M31 = |■8(0&0@3&0)| = 0(0) – 3(0) = 0 M32 = |■8(1&0@3&0)| = 1(0) – 3(0) = 0 M33 = |■8(1&0@3&3)| = 1(3) – 3(0) = 3 A11 = ( – 1)1 + 1 M11 = ( – 1)2 (– 3) = 1 ( – 3) = – 3 A12 = ( – 1)1+2 M12 = ( – 1)3 ( – 3) = ( – 1) (– 3) = 3 A13 = ( – 1)1+3 M13 = ( – 1)4 ( – 9) = 1 . (– 9) = – 9 A21 = ( – 1)2+1 M21 = ( – 1)3 (0) = 0 A22 = ( – 1)2+2 M22 = ( – 1)4 ( – 1) = 1 . ( – 1) = −1 A23 = ( – 1)2+3 M23 = ( – 1)5 (2) = – 1 (2) = – 2 A31 = ( – 1)3+1 M31 = ( – 1)4 (2) = 0 A32 = ( – 1)3+2 M32 = ( – 1)5 (0) = 0 A33 = ( – 1)3+3 M33 = ( – 1)6 3 = 3 ∴ adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(−3&0&0@3&−1&0@−9&−2&3)] Step 3: Calculate A–1 A– 1 = 1/(|A|) ( adj (A)) = 1/(−3) [■8(−3&0&0@3&−1&0@−9&−2&3)] = (−𝟏)/𝟑 [■8(−𝟑&𝟎&𝟎@𝟑&−𝟏&𝟎@−𝟗&−𝟐&𝟑)]

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