Example 28 - Chapter 4 Class 12 Determinants

Transcript

Hence A = 3﷮−2﷮3﷮2﷮1﷮−1﷮4﷮−3﷮2﷯﷯, 𝑥= 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 8﷮1﷮4﷯﷯ Step 2 Calculate |A| |A| = 3﷮−2﷮3﷮2﷮1﷮−1﷮4﷮−3﷮2﷯﷯ = 3 1﷮−1﷮−3﷮2﷯﷯ – 1( – 2) 2﷮−1﷮4﷮2﷯﷯ + 3 2﷮1﷮4﷮−3﷯﷯ = 3 (2 – 3 ) – 2 (4 + 4) + 3 ( – 6 – 4) = 3 ( – 1) + 2(8) + 3 ( – 10) = – 3 + 16 – 30 = – 17 Thus, |A| ≠ 0 ∴ 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﷮13﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮13﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯ A = 3﷮−2﷮3﷮2﷮1﷮−1﷮4﷮−3﷮2﷯﷯ M11 = 1﷮−1﷮−3﷮2﷯﷯ = 2 – 3 = – 1 M12 = 2﷮−1﷮4﷮2﷯﷯ = 4 + 4 = 8 M13 = 2﷮1﷮4﷮−3﷯﷯ = – 6 – 4 = – 10 M21 = −2﷮3﷮−3﷮2﷯﷯ = – 4 + 9 = 5 M22 = 3﷮3﷮4﷮2﷯﷯ = 6 – 12 = – 6 M23 = 3﷮−2﷮4﷮−3﷯﷯ = – 9 + 8 = – 1 M31 = −2﷮3﷮1﷮−1﷯﷯ = 1 – 2 = – 1 M32 = 3﷮3﷮2﷮−1﷯﷯ = – 3 – 6 = – 9 M33 = 3﷮−2﷮2﷮1﷯﷯ = 3 + 4 = 7 A11 = ( – 1)1+1 . M11 = ( –1 )2 . ( – 1) = – 1 A12 = ( – 1)1+2 . M12 = ( –1 )3 . 8 = – 8 A13 = ( – 1)1+3 . M13 = ( –1 )4 . ( – 10) = – 10 A21 = ( – 1)2+1 . M21 = ( –1 )3 . (5) = – 5 A22 = ( – 1)2+2 . M22 = ( –1 )4 . ( – 6) = – 6 A23 = ( – 1)2+3 . ( – 1) = ( –1 )5 . ( – 1) = 1 A31 = ( – 1)3+1 . M31 = ( –1 )4 . ( – 1) = – 1 A32 = ( – 1)3+2 . M32 = ( –1 )5 . (– 9) = 9 A33 = ( – 1)3+3 . M33 = ( –1 )6 . 7 = 7 Thus, adj (A) = −1﷮−5﷮−1﷮−8﷮−6﷮9﷮−10﷮1﷮7﷯﷯ Now, A-1 = 1﷮|A|﷯ adj A Putting values = 1﷮−17﷯ −1﷮−5﷮−1﷮−8﷮−6﷮9﷮−10﷮1﷮7﷯﷯ Also, X = A-1 B Putting values 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮−17﷯ −1﷮−5﷮−1﷮−8﷮−6﷮9﷮−10﷮1﷮7﷯﷯ 8﷮1﷯﷮4﷯﷯ = 1﷮−17﷯ −1 8﷯+ −5﷯ 1﷯+ −1﷯4﷮−8 8﷯+ −6﷯ 1﷯+9 4﷯﷮−10 8﷯+1 1﷯+7 4﷯﷯﷯ 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮−17﷯ −8﷮−5﷮−5﷮−64﷮−6﷮+36﷮−80﷮+1﷮+28﷯﷯ = 1﷮−17﷯ −17﷮−36﷯﷮−51﷯﷯ = −17﷮−17﷯﷮ −34﷮−(17)﷯﷯﷮ −51﷮−17﷯﷯﷯ 𝑥﷮𝑦﷯﷮𝑧﷯﷯ = 1﷮2﷯﷮3﷯﷯ Hence x = 1, y = 𝟐 & z = 3