Check sibling questions

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

Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 6
Ex 4.6, 13 - Chapter 4 Class 12 Determinants - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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 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 M22 = |■8(2&3@3&−2)| = –4 – 9 = −13 M23 = |■8(2&3@3&−1)| = –2 – 9 = –11 M31 = |■8(3&3@−2&1)| = 3 + 6 = 9 M32 = |■8(2&3@1&1)| = 2 − 3 = – 1 M33 = |■8(2&3@1&−2)| = –4 − 3 = –7 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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.