Ex 4.6, 13 - Solve linear equations using matrix method - Ex 4.6

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Ex 4.6, 13 Solve system of linear equations, using matrix method. 2x + 3y + 3z = 5 x − 2y + z = −4 3x − y − 2z = 3 The system of equations is 2x + 3y + 3z = 5 x − 2y + z = −4 3x − y − 2z = 3 Step 1 Write equation as AX = B 2﷮3﷮3﷮1﷮−2﷮1﷮3﷮−1﷮−2﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 5﷮−4﷮3﷯﷯ Hence A = 2﷮3﷮3﷮1﷮−2﷮1﷮3﷮−1﷮−2﷯﷯ × = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 5﷮−4﷮3﷯﷯ Step 2 Calculate |A| |A| = 2﷮3﷮3﷮1﷮−2﷮1﷮3﷮−1﷮−2﷯﷯ = 2 −2﷮1﷮−1﷮−2﷯﷯ – 3 1﷮1﷮3﷮−2﷯﷯ + 2 1﷮−2﷮3﷮−1﷯﷯ = 2 ( 4 + 1) – 3 ( – 2 – 3) + 3 ( – 2 + 6) = 2 (5) – 3 ( – 5) + 3 (5) = 10 + 15 + 5 = 40 Since |A|≠ 0 ∴ The system of equations is consistent & has a unique solution 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 = 2﷮3﷮3﷮1﷮−2﷮1﷮3﷮−1﷮−2﷯﷯ M11 = −2﷮1﷮−1﷮−2﷯﷯ = 4 + 1 = 5 M12 = 1﷮1﷮3﷮−2﷯﷯ = – 2 – 3 = – 5 M13 = 1﷮−2﷮3﷮−1﷯﷯ = – 1 + 6 = 5 M21 = 3﷮3﷮−1﷮−2﷯﷯ = − 6 + 3 = – 3 M22 = 2﷮3﷮3﷮−2﷯﷯ = – 1 – 9 = − 13 M23 = 2﷮3﷮3﷮−1﷯﷯ = – 2 – 9 = – 11 M31 = 3﷮3﷮−2﷮1﷯﷯ = 3 + 6 = 9 M32 = 2﷮3﷮1﷮1﷯﷯ = 2 – 3 = 1 M33 = 2﷮3﷮1﷮−2﷯﷯ = – 4 + 3 = – 7 A11 = ( –1)﷮1+1﷯ M11 = ( –1)2 . (5) = 5 A12 = ( –1)﷮1+2﷯ M12 = ( –1)﷮3﷯ . ( – 5) = 5 A13 = ( −1)﷮1+3﷯ M13 = ( −1)﷮4﷯ . (5) = 5 A21 = ( −1)﷮2+1﷯ M21 = ( −1)﷮3﷯ . ( – 3) = 3 A22 = ( −1)﷮2+2﷯ M22 = ( –1)4 . ( – 13) = – 13 A23 = (−1)﷮2+3﷯. M23 = (−1)﷮5﷯. ( – 11) = 11 A31 = (−1)﷮3+1﷯. M31 = (−1)﷮4﷯ . (9) = 9 A32 = (−1)﷮3+2﷯ . M32 = (−1)﷮5﷯. ( – 1) = 1 A33 = (−1)﷮3+3﷯ . M33 = ( –1)6 . ( – 7) = –7 Thus, adj A = 5﷮3﷮9﷮5﷮−13﷮1﷮5﷮11﷮−7﷯﷯ & |A| = 40 So, A-1 = 1﷮|A|﷯ adj A A-1 = 1﷮40﷯ 5﷮3﷮9﷮5﷮−13﷮1﷮5﷮11﷮−7﷯﷯ & B = 5﷮−4﷮3﷯﷯ Now, solving X = A-1 B 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮40﷯ 5﷮3﷮9﷮5﷮−13﷮1﷮5﷮11﷮−7﷯﷯ 5﷮−4﷮3﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮40﷯ 5 5﷯+3 −4﷯+9(3)﷮5 5﷯+(−13) −4﷯+1(3)﷮5 5﷯+11 −4﷯+(−7)(3)﷯﷯ = 1﷮40﷯ 25−12+27﷮25+52+3﷮25−44−21﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮40﷯ 25−12+27﷮25+52+3﷮25−44−21﷯﷯ = 1﷮40﷯ 40﷮80﷮−40﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮2﷮−1﷯﷯ Hence x = 1 , y = 2, & z = –1 But X = 𝑥﷮𝑦﷮𝑧﷯﷯ So, 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮2﷮−1﷯﷯ Hence x = 1 , y = 2, & z = –1

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