Check sibling questions

Ex 4.5, 8 - Find inverse of matrix [1 0 0 3 3 0 5 2 -1] - Ex 4.5

Ex 4.5, 8 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 8 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.5, 8 - Chapter 4 Class 12 Determinants - Part 4

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Transcript

Ex 4.5, 8 Find the inverse of each of the matrices (if it exists). 1﷮0﷮0﷮3﷮3﷮0﷮5﷮2﷮−1﷯﷯ Let A = 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| = 1﷮0﷮0﷮3﷮3﷮0﷮5﷮2﷮−1﷯﷯ = 1 3﷮0﷮2﷮−1﷯﷯ – 0 3﷮0﷮5﷮1﷯﷯ + 0 3﷮3﷮5﷮2﷯﷯ = 1(–3 – 0) + 0 + 0 = −3 Since |A| ≠ 0 , A–1 exists Step 2: Calculate adj A adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ A = 1﷮0﷮0﷮3﷮3﷮0﷮5﷮2﷮−1﷯﷯ M11 = 3﷮0﷮2﷮−1﷯﷯ = 3(−1) – 2(0) = −3 M12 = 3﷮0﷮5﷮−1﷯﷯ = 3(–1) – 0(5) = – 3 M13 = 3﷮3﷮5﷮2﷯﷯ = 3(2) – 5(3) = – 9 M21 = 0﷮0﷮2﷮–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) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ = −3﷮0﷮0﷮3﷮−1﷮0﷮−9﷮−2﷮3﷯﷯ Step 3: Calculate A–1 A– 1 = 1﷮|A|﷯ ( adj (A)) = 1﷮−3﷯ −3﷮0﷮0﷮3﷮−1﷮0﷮−9﷮−2﷮3﷯﷯ = −𝟏﷮𝟑﷯ −𝟑﷮𝟎﷮𝟎﷮𝟑﷮−𝟏﷮𝟎﷮−𝟗﷮−𝟐﷮𝟑﷯﷯

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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.