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Ex 4.5
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Last updated at March 16, 2023 by Teachoo
Ex 4.5, 8 Find the inverse of each of the matrices (if it exists). 10033052−1 Let A = 10033052−1 We know that A–1 = 1|A| (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = 10033052−1 = 1 302−1 – 0 3051 + 0 3352 = 1(–3 – 0) + 0 + 0 = −3 Since |A| ≠ 0 , A–1 exists Step 2: Calculate adj A adj (A) = A11A21A31A12A22A32A33A23A33 A = 10033052−1 M11 = 302−1 = 3(−1) – 2(0) = −3 M12 = 305−1 = 3(–1) – 0(5) = – 3 M13 = 3352 = 3(2) – 5(3) = – 9 M21 = 002–1 = 0(–1) – 2(0) = 0 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) = A11A21A31A12A22A32A33A23A33 = −3003−10−9−23 Step 3: Calculate A–1 A– 1 = 1|A| ( adj (A)) = 1−3 −3003−10−9−23 = −𝟏𝟑 −𝟑𝟎𝟎𝟑−𝟏𝟎−𝟗−𝟐𝟑