Ex 4.6, 14 - Chapter 4 Class 12 Determinants

Transcript

Ex 4.6, 14 Solve system of linear equations, using matrix method. x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 The system of equations are x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 Step 1 Write equation as AX = B 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 7﷮−5﷮12﷯﷯ Hence A = 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ , X = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 7﷮−5﷮12﷯﷯ Step 2 Calculate |A| |A| = 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ = 1 4﷮−5﷮−1﷮−3﷯﷯ – ( –1) 3﷮−5﷮2﷮3﷯﷯ + 2 3﷮4﷮2﷮−1﷯﷯ = 1 ( 12 – 5) + 1 (9 + 10) + 2 ( – 3 – 8) = 2 (7) + 1 (19) + 2 ( – 11) = 7 + 19 – 22 = 4 |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1﷮|A|﷯ adj (A) adj A = A11﷮A12﷮A13﷮A21﷮A22﷮A23﷮A31﷮A32﷮A33﷯﷯﷮′﷯ = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A13﷮A23﷮A33﷯﷯ A = 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ M11 = 4﷮−5﷮−1﷮3﷯﷯ = 12 – 5 = 7 M12 = 3﷮−5﷮2﷮3﷯﷯ = 9 + 10 = 19 M13 = 3﷮4﷮2﷮−1﷯﷯ = – 3 – 8 = – 11 M21 = −1﷮2﷮−1﷮3﷯﷯ = − 3 + 2 = – 1 M22 = 1﷮2﷮2﷮3﷯﷯ = 3 – 4 = – 1 M23 = 1﷮−1﷮2﷮−1﷯﷯ = – 1 + 2 = 1 M31 = −1﷮2﷮4﷮−5﷯﷯ = 5 – 8 = – 3 M32 = 1﷮2﷮3﷮−5﷯﷯ = – 5 – 6 = – 11 M33 = 1﷮−1﷮3﷮4﷯﷯ = 4 + 3= 7 A11 = ( –1)﷮1+1﷯ M11 = ( –1)2 . 7 = 7 A12 = ( –1)﷮1+2﷯ M12 = ( –1)﷮3﷯ . 19 = – 19 A13 = ( −1)﷮1+3﷯ . M13 = ( −1)﷮4﷯ . – 11= – 11 A21 = ( −1)﷮2+1﷯ M21 = ( −1)﷮3﷯ . ( – 1) = + 1 A22 = ( −1)﷮2+2﷯ M22 = ( –1)4 . ( – 1) = – 1 A23 = (−1)﷮2+3﷯. M23 = (−1)﷮5﷯. 1 = –1 A31 = (−1)﷮3+1﷯. M31 = (−1)﷮4﷯ . ( – 3) = – 3 A32 = (−1)﷮3+2﷯ . M32 = (−1)﷮5﷯. ( – 11) = 11 A33 = (−1)﷮3+3﷯ . M33 = ( –1)6 . ( – 7) = 7 Thus adj A = 7﷮1﷮−3﷮−19﷮−1﷮11﷮−11﷮−1﷮7﷯﷯ & |A| = 4 Now, A-1 = 1﷮|A|﷯ adj A A-1 = 1﷮4﷯ 7﷮1﷮−3﷮−19﷮−1﷮11﷮−11﷮−1﷮7﷯﷯ & B = 7﷮−5﷮12﷯﷯ Solving X = A-1 B 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮4﷯ 7﷮1﷮−3﷮−19﷮−1﷮11﷮−11﷮−1﷮7﷯﷯ 7﷮−5﷮12﷯﷯ = 1﷮4﷯ 7 7﷯+1 −5﷯+(−13)(12)﷮−19 7﷯+(−1) −4﷯+11(12)﷮−11 7﷯+1 −5﷯+7(12)﷯﷯ = 1﷮4﷯ 49−5−36﷮−133+5+132﷮−77+5+84﷯﷯ = 1﷮4﷯ 8﷮4﷮12﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 2﷮1﷮3﷯﷯ Hence x = 2 , y = 1, & z = 3