Ex 4.4, 8 - Chapter 4 Class 12 Determinants

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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)]