Ex 4.5, 10 - Class 12

Transcript

Ex 4.5, 10 Find the inverse of each of the matrices (if it exists). 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ Let A = 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ We know that A–1 = 1﷮|A|﷯ (adj A) exists if |A|≠ 0 Step 1: Calculating |A| |A| = 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ = 1 2﷮−3﷮−2﷮4﷯﷯– (– 1) 0﷮−3﷮3﷮4﷯﷯ + 2 0﷮2﷮3﷮−2﷯﷯ = 1(8 – 6) + 1 (0 + 9) + 2 (0 – 6) = 1 (2) + 1 (9) + 2 ( – 6) = – 1 Since |A| ≠ 0, A–1 exists Step 3: Calculate (adj A) adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ A = 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ M11 = 2﷮−3﷮−2﷮4﷯﷯=2(4)–(-2)(-3)= 2 M12 = 0﷮−3﷮3﷮4﷯﷯ = 0(4) – 3(-3) = 9 M13 = 0﷮2﷮3﷮−2﷯﷯= 0(-2) – 3(2) = –6 M21 = −1﷮2﷮−2﷮4﷯﷯=(–1)(4)–(-2)(2) = 0 A11 = ( – 1)1 + 1 M11 = ( – 1)2 2 = 2 A12 = ( – 1)1+2 M12 = ( – 1)3 9 = – 9 A13 = ( – 1)1+3 M13 = ( – 1)4 (–6) = –6 A21 = ( – 1)2+1 M21 = ( – 1)3 0 = 0 A22 = ( – 1)2+2 M22 = ( – 1)4 ( –2) = – 2 A23 = ( – 1)2+3 M23 = ( – 1)5 (1) = (– 1) (1) = (– 1) A31 = ( – 1)3+1 M31 = ( – 1)4 (– 1) = 1 ( – 1) = – 1 A32 = ( – 1)3+2 M32 = ( – 1)5 (– 3) = ( – 1) ( – 3) = 3 A33 = ( – 1)3+3 M33 = ( – 1)6 ( 2) = 2 adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ = 2﷮0﷮−1﷮−9﷮−2﷮3﷮−6﷮−1﷮2﷯﷯ Step 3: Calculate inverse A– 1 = 1﷮|A|﷯ ( adj (A)) = 1﷮−1﷯ 2﷮0﷮−1﷮−9﷮−2﷮3﷮−6﷮−1﷮2﷯﷯ = – 2﷮0﷮−1﷮−9﷮−2﷮3﷮−6﷮−1﷮2﷯﷯ = −𝟐﷮𝟎﷮𝟏﷮𝟗﷮𝟐﷮−𝟑﷮𝟔﷮𝟏﷮−𝟐﷯﷯