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Ex 4.5, 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 Writing equation as AX = B [■8(2&3&3@1&−2&1@3&−1&−2)] [■8(𝑥@𝑦@𝑧)] = [■8(5@−4@3)] Hence, A = [■8(2&3&3@1&−2&1@3&−1&−2)] × = [■8(𝑥@𝑦@𝑧)] & B = [■8(5@−4@3)] Calculating |A| |A| = |■8(2&3&3@1&−2&1@3&−1&−2)| = 2 |■8(−2&1@−1&−2)| – 3 |■8(1&1@3&−2)| + 3 |■8(1&−2@3&−1)| = 2 (4 + 1) – 3 (−2 − 3) + 3 (–1 + 6) = 2 (5) – 3 (–5) + 3 (5) = 10 + 15 + 15 = 40 Since |A|≠ 0 ∴ The system of equations is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj A = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(2&3&3@1&−2&1@3&−1&−2)] M11 = [■8(−2&1@−1&−2)] = 4 + 1 = 5 M12 = |■8(1&1@3&−2)| = –2 – 3 = – 5 M13 = |■8(1&−2@3&−1)| = –1 + 6 = 5 M21 = |■8(3&3@−1&−2)| = −6 + 3 = – 3 Now, 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 = [■8(5&3&9@5&−13&1@5&11&−7)] Now, A-1 = 1/(|A|) adj A A-1 = 1/40 [■8(5&3&9@5&−13&1@5&11&−7)] Also, X = A-1 B [■8(𝑥@𝑦@𝑧)] = 1/40 [■8(5&3&9@5&−13&1@5&11&−7)] [■8(5@−4@3)] [■8(𝑥@𝑦@𝑧)] = 1/40 [■8(5(5)+3(−4)+9(3)@5(5)+(−13)(−4)+1(3)@5(5)+11(−4)+(−7)(3))] = 1/40 [■8(25−12+27@25+52+3@25−44−21)] [■8(𝑥@𝑦@𝑧)] = 1/40 [■8(25−12+27@25+52+3@25−44−21)] = 1/40 [■8(40@80@−40)] [■8(𝑥@𝑦@𝑧)] = [■8(1@2@−1)] Hence, x = 1 , y = 2, & z = –1

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo