Ex 4.6

Chapter 4 Class 12 Determinants
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Ex 4.6, 14 Solve system of linear equations, using matrix method. x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 The system of equations are x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 Writing equation as AX = B [■8(1&−1&[email protected]&4&−[email protected]&−1&3)] [■8(𝑥@𝑦@𝑧)] = [■8([email protected][email protected])] Hence, A = [■8(1&−1&[email protected]&4&−[email protected]&−1&3)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8([email protected][email protected])] Calculating |A| |A| = |■8(1&−1&[email protected]&4&−[email protected]&−1&3)| = 1 |■8(4&−[email protected]−1&−3)| – ( –1) |■8(3&−[email protected]&3)| + 2 |■8(3&[email protected]&−1)| = 1 ( 12 – 5) + 1 (9 + 10) + 2 (–3 – 8) = 2 (7) + 1 (19) + 2 (–11) = 7 + 19 – 22 = 4 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj A = [■8(A11&A12&[email protected]&A22&[email protected]&A32&A33)]^′ = [■8(A11&A21&[email protected]&A22&[email protected]&A23&A33)] A = [■8(1&−1&[email protected]&4&−[email protected]&−1&3)] M11 = [■8(4&−[email protected]−1&3)] = 12 – 5 = 7 M12 = |■8(3&−[email protected]&3)| = 9 + 10 = 19 M13 = |■8(3&[email protected]&−1)| = –3 – 8 = –11 M21 = |■8(−1&[email protected]−1&3)| = −3 + 2 = –1 M22 = |■8(1&[email protected]&3)| = 3 – 4 = –1 M23 = |■8(1&−[email protected]&−1)| = –1 + 2 = 1 M31 = |■8(−1&[email protected]&−5)| = 5 – 8 = –3 M32 = |■8(1&[email protected]&−5)| = –5 – 6 = –11 M33 = |■8(1&−[email protected]&4)| = 4 + 3= 7 Now, A11 = 〖"(–1)" 〗^(1+1) M11 = (–1)2 . 7 = 7 A12 = 〖"(–1)" 〗^"1+2" M12 = 〖"(–1)" 〗^"3" . 19 = – 19 A13 = 〖(−1)〗^(1+3) . M13 = 〖(−1)〗^4 . – 11= –11 A21 = 〖(−1)〗^(2+1) M21 = 〖(−1)〗^3 . (–1) = + 1 A22 = 〖(−1)〗^(2+2) M22 = (–1)4 . (–1) = –1 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. 1 = –1 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . (–3) = –3 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. (–11) = 11 A33 = 〖(−1)〗^(3+3) . M33 = (–1)6 . (–7) = 7 Thus, adj A = [■8(7&1&−[email protected]−19&−1&[email protected]−11&−1&7)] Now, A-1 = 1/(|A|) adj A A-1 = 1/4 [■8(7&1&−[email protected]−19&−1&[email protected]−11&−1&7)] Solving X = A-1 B [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(7&1&−[email protected]−19&−1&[email protected]−11&−1&7)] [■8([email protected][email protected])] [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(7(7)+1(−5)+(−13)(12)@−19(7)+(−1)(−4)+11(12)@−11(7)+1(−5)+7(12))] [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(49−5−[email protected][email protected]−77+5+84)] = 1/4 [■8([email protected]@12)] [■8(𝑥@𝑦@𝑧)] = [■8([email protected]@3)] Hence, x = 2 , y = 1, & z = 3

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.