1. Chapter 4 Class 12 Determinants
2. Serial order wise

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Step 2 Calculate |A| |A| = 2﷮1﷮1﷮1﷮−2﷮−1﷮0﷮3﷮−5﷯﷯ = 2 −2﷮−1﷮3﷮−5﷯﷯ – 1 1﷮−1﷮0﷮−5﷯﷯ + 1 1﷮−2﷮0﷮3﷯﷯ = 2 (10 + 3 ) – 1 (– 5 + 0) + 1 (3 – 0) = 2 (13) – 1 (– 5 ) + 21 (3) = 34 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate X = A-1 B A-1 = 1﷮|A|﷯ adj (A) adj (A) = A﷮11﷯﷮ A﷮12﷯﷮ A﷮13﷯﷮ A﷮21﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮31﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯﷮′﷯ = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮13﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯ A = 2﷮1﷮1﷮1﷮−2﷮−1﷮0﷮3﷮−5﷯﷯ M11 = −2﷮−1﷮3﷮−5﷯﷯ = 15 + 3 = 13 M12 = 1﷮−1﷮0﷮−5﷯﷯ = – 5 + 0 = – 5 M13 = 1﷮−2﷮0﷮3﷯﷯ = 3 + 0 = 3 M21 = 1﷮1﷮3﷮−5﷯﷯ = – 5 – 3 = – 8 M22 = 2﷮1﷮0﷮−5﷯﷯ = – 10 + 0 = – 10 M23 = 2﷮1﷮0﷮3﷯﷯ = 6 + 0 = 6 M31 = 1﷮1﷮−2﷮−1﷯﷯ = – 1 + 2 = 1 M32 = 2﷮1﷮1﷮−1﷯﷯ = – 2 – 1 = – 3 M33 = 2﷮1﷮1﷮−2﷯﷯ = – 4 – 1 = – 5 A11 = ( – 1)1+1 . M11 = ( –1 )2 . (13) = 13 A12 = ( – 1)1+2 . M12 = ( –1 )3 . (– 5) = 5 A13 = ( – 1)1+3 . M13 = ( –1 )4 . (3) = 3 A21 = ( – 1)2+1 . M21 = ( –1 )3 . ( – 8) = 8 A22 = ( – 1)2+2 . M22 = ( –1 )4 . ( – 10) = – 10 A23 = ( – 1)2+3 . M23 = ( –1 )5 . (6) = – 6 A31 = ( – 1)3+1 . M31 = ( –1 )4 . (1) = 1 A32 = ( – 1)3+2 . M32 = ( –1 )5 . (– 3) = 3 A33 = ( – 1)3+3 . M33 = ( –1 )6 . (– 5) = – 5 Thus, adj (A) = 13﷮8﷮1﷮5﷮−10﷮3﷮3﷮−6﷮−5﷯﷯ Now, A-1 = 1﷮|A|﷯ adj A Putting values = 1﷮34﷯ 13﷮8﷮1﷮5﷮−10﷮3﷮3﷮−6﷮−5﷯﷯ Also, X = A-1 B Putting values 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮34﷯ 13﷮8﷮1﷮5﷮−10﷮3﷮3﷮−6﷮−5﷯﷯ 1﷮ 3﷮2﷯﷯﷮9﷯﷯ 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮34﷯ 13 1﷯+8 3﷮2﷯﷯+1 9﷯﷮5 1﷯+ −10﷯ 3﷮2﷯﷯+3 9﷯﷮3 1﷯+ −6﷯ 3﷮2﷯﷯+ −5﷯ 9﷯﷯﷯ 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮34﷯ 13+12+9﷮5−15+27﷮3−9−45﷯﷯ = 1﷮34﷯ 34﷮17﷯﷮−51﷯﷯ 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮ 1﷮2﷯﷯﷮ −3﷮2﷯﷯﷯ Hence x = 1, y = 𝟏﷮𝟐﷯ & z = −𝟑﷮𝟐﷯