Ex 4.5, 13 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Last updated at April 16, 2024 by Teachoo
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