Ex 4.6, 12 - Class 12

Transcript

Ex 4.6, 12 Solve system of linear equations, using matrix method. x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 The system of equations is x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 Step 1 Write equation as AX = B 1﷮−1﷮1﷮2﷮1﷮−3﷮1﷮1﷮1﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 4﷮0﷮2﷯﷯ Hence A = 1﷮−1﷮1﷮2﷮1﷮−3﷮1﷮1﷮1﷯﷯ , X = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 4﷮0﷮2﷯﷯ Step 2 Calculate |A| |A| = 1﷮−1﷮1﷮2﷮1﷮−3﷮1﷮1﷮1﷯﷯ = 1 1﷮−3﷮1﷮1﷯﷯ – ( –1) 2﷮−3﷮1﷮1﷯﷯ + 1 2﷮1﷮1﷮1﷯﷯ = ( 1 + 3) + 1 ( 2 + 3) + 1 (2 – 1) = 1 (4) + 1 (5) + 1 (1) = 4 + 5 + 1 = 10 Since |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 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﷮1﷮2﷮1﷮−3﷮1﷮1﷮1﷯﷯ M11 = 1﷮−3﷮1﷮1﷯﷯ = 1 + 3 = 4 M12 = 2﷮−3﷮1﷮1﷯﷯ = 2 + 3 = 5 M13 = 2﷮1﷮1﷮1﷯﷯ = 2 – 1= 1 M21 = −1﷮1﷮1﷮1﷯﷯ = − 1 – 1 = – 2 M22 = 1﷮1﷮1﷮1﷯﷯ = 1 – 1 = 0 M23 = 1﷮−1﷮1﷮1﷯﷯ = 1 + 1 = 2 M31 = −1﷮1﷮1﷮−3﷯﷯ = 3 – 1 = 2 M32 = 1﷮1﷮2﷮−3﷯﷯ = – 3 – 4 = – 5 M33 = 1﷮−1﷮2﷮1﷯﷯ = 3 + 2 = 3 A11 = ( –1)﷮1+1﷯ M11= ( –1)2 . 4= 4 A12 = ( –1)﷮1+2﷯ M12 = ( –1)﷮3﷯ . 5 = – 5 A13 = ( −1)﷮1+3﷯ M13= ( −1)﷮4﷯ . (1) = 1 A21 = −1﷯﷮2+1﷯ M21= ( −1)﷮3﷯ . (-2) = 2 A22 = ( −1)﷮2+2﷯ M22 = ( –1)4 . 0 = 0 A23 = (−1)﷮2+3﷯. M23 = (−1)﷮5﷯. ( 2) = – 2 A31 = (−1)﷮3+1﷯. M31 = (−1)﷮4﷯ . (2) = 2 A32 = (−1)﷮3+2﷯ . M32 = (−1)﷮5﷯. ( – 5) = 5 A33 = (−1)﷮3+3﷯ . M33 = ( –1)6 . 3 = 3 Thus , adj A = 4﷮2﷮2﷮−5﷮0﷮5﷮1﷮−2﷮3﷯﷯ & |A| = 10 So, A-1 = 1﷮|A|﷯ adj A A-1 = 1﷮10﷯ 4﷮2﷮2﷮−5﷮0﷮5﷮1﷮−2﷮3﷯﷯ & B = 4﷮0﷮2﷯﷯ Now, solving X = A-1 B 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮10﷯ 4﷮2﷮2﷮−5﷮0﷮5﷮1﷮1﷮3﷯﷯ 4﷮0﷮2﷯﷯ = 1﷮10﷯ 4 4﷯+2 0﷯+2(2)﷮0 4﷯+0 0﷯+5(2)﷮2 4﷯+1 0﷯+3(2)﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮10﷯ 16+0+4﷮−20+0+10﷮4+0+6﷯﷯ = 1﷮10﷯ 20﷮−10﷮10﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 2﷮−1﷮1﷯﷯ Hence, x = 2 , y = –1, & z = 1