# Ex 4.6, 12 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.6, 12 Solve system of linear equations, using matrix method. x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 The system of equations is x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 Step 1 Write equation as AX = B 1−1121−3111 𝑥𝑦𝑧 = 402 Hence A = 1−1121−3111 , X = 𝑥𝑦𝑧 & B = 402 Step 2 Calculate |A| |A| = 1−1121−3111 = 1 1−311 – ( –1) 2−311 + 1 2111 = ( 1 + 3) + 1 ( 2 + 3) + 1 (2 – 1) = 1 (4) + 1 (5) + 1 (1) = 4 + 5 + 1 = 10 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate X = A-1 B Calculating A-1 Now, A-1 = 1|A| adj (A) adj A = A11A12A13A21A22A23A31A32A33′ = A11A21A31A12A22A32A13A23A33 A = 1−1121−3111 M11 = 1−311 = 1 + 3 = 4 M12 = 2−311 = 2 + 3 = 5 M13 = 2111 = 2 – 1= 1 M21 = −1111 = − 1 – 1 = – 2 M22 = 1111 = 1 – 1 = 0 M23 = 1−111 = 1 + 1 = 2 M31 = −111−3 = 3 – 1 = 2 M32 = 112−3 = – 3 – 4 = – 5 M33 = 1−121 = 3 + 2 = 3 A11 = ( –1)1+1 M11= ( –1)2 . 4= 4 A12 = ( –1)1+2 M12 = ( –1)3 . 5 = – 5 A13 = ( −1)1+3 M13= ( −1)4 . (1) = 1 A21 = −12+1 M21= ( −1)3 . (-2) = 2 A22 = ( −1)2+2 M22 = ( –1)4 . 0 = 0 A23 = (−1)2+3. M23 = (−1)5. ( 2) = – 2 A31 = (−1)3+1. M31 = (−1)4 . (2) = 2 A32 = (−1)3+2 . M32 = (−1)5. ( – 5) = 5 A33 = (−1)3+3 . M33 = ( –1)6 . 3 = 3 Thus , adj A = 422−5051−23 & |A| = 10 So, A-1 = 1|A| adj A A-1 = 110 422−5051−23 & B = 402 Now, solving X = A-1 B 𝑥𝑦𝑧 = 110 422−505113 402 = 110 4 4+2 0+2(2)0 4+0 0+5(2)2 4+1 0+3(2) 𝑥𝑦𝑧 = 110 16+0+4−20+0+104+0+6 = 110 20−1010 𝑥𝑦𝑧 = 2−11 Hence, x = 2 , y = –1, & z = 1

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.