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 Ex 4.5, 9 - Find inverse of [2 1 3 4 -1 0 -7 2 1] - Class 12 - Ex 4.5

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.5, 9 Find the inverse of each of the matrices (if it exists). 2﷮1﷮3﷮4﷮−1﷮0﷮−7﷮2﷮1﷯﷯ Let A = 2﷮1﷮3﷮4﷮−1﷮0﷮−7﷮2﷮1﷯﷯ We know that A–1 = 1﷮|A|﷯ (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = 2﷮1﷮3﷮4﷮−1﷮0﷮−7﷮2﷮1﷯﷯ = 2 –1﷮0﷮2﷮1﷯﷯ – 1 4﷮0﷮−7﷮1﷯﷯ + 3 4﷮−1﷮−7﷮2﷯﷯ = 2(–1 – 0) – 1 (4 – 0) + 3 (8 – 7) = 2 ( – 1) – 1 (4) + 3 (1) = – 3 Since 𝐴﷯ ≠ 0 , A–1 exists Step 2: Calculating adj (A) adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ A = 2﷮1﷮3﷮4﷮−1﷮0﷮−7﷮2﷮1﷯﷯ M11 = −1﷮0﷮2﷮1﷯﷯ = –1(1) – 0(2) = -1 M12 = 4﷮0﷮−7﷮1﷯﷯ = 4(1) – 0(–7) = 4 M13 = 4﷮−1﷮−7﷮2﷯﷯ = 4(2) – (-7)(-1) = 1 M21 = 1﷮3﷮2﷮1﷯﷯ = 1(1) – 2(3) = – 5 A11 = ( – 1)1 + 1 M11 = ( – 1)2 (– 1) = 1 ( – 1) = – 1 A12 = ( – 1)1+2 M12 = ( – 1)3 ( 4) = ( – 1) (4) = –4 A13 = ( – 1)1+3 M13 = ( – 1)4 (1) = 1 A21 = ( – 1)2+1 M21 = ( – 1)3 ( –5) = (−1)(−5) = 5 A22 = ( – 1)2+2 M22 = ( – 1)4 ( –23) = 23 A23 = ( – 1)2+3 M23 = ( – 1)5 11 = −1 (11) = – 11 A31 = ( – 1)3+1 M31 = ( – 1)4 (4) = 1 (3) = 3 A32 = ( – 1)3+2 M32 = ( – 1)5 (– 12) = ( – 1) ( – 12) = 12 A33 = ( – 1)3+3 M33 = ( – 1)6 ( – 6) = 1 . ( – 6) = – 6 Thus, adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ = −1﷮5﷮3﷮−4﷮23﷮12﷮1﷮−11﷮−6﷯﷯ Step 3: Calculate inverse A– 1 = 1﷮|A|﷯ ( adj (A)) = 1﷮−3﷯ −1﷮5﷮3﷮−4﷮23﷮12﷮1﷮−11﷮−6﷯﷯ = −𝟏﷮𝟑﷯ −𝟏﷮𝟓﷮𝟑﷮−𝟒﷮𝟐𝟑﷮𝟏𝟐﷮𝟏﷮−𝟏𝟏﷮−𝟔﷯﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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