# Ex 4.6, 14

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 Step 1 Write equation as AX = B 1−1234−52−13 𝑥𝑦𝑧 = 7−512 Hence A = 1−1234−52−13 , X = 𝑥𝑦𝑧 & B = 7−512 Step 2 Calculate |A| |A| = 1−1234−52−13 = 1 4−5−1−3 – ( –1) 3−523 + 2 342−1 = 1 ( 12 – 5) + 1 (9 + 10) + 2 ( – 3 – 8) = 2 (7) + 1 (19) + 2 ( – 11) = 7 + 19 – 22 = 4 |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1|A| adj (A) adj A = A11A12A13A21A22A23A31A32A33′ = A11A21A31A12A22A32A13A23A33 A = 1−1234−52−13 M11 = 4−5−13 = 12 – 5 = 7 M12 = 3−523 = 9 + 10 = 19 M13 = 342−1 = – 3 – 8 = – 11 M21 = −12−13 = − 3 + 2 = – 1 M22 = 1223 = 3 – 4 = – 1 M23 = 1−12−1 = – 1 + 2 = 1 M31 = −124−5 = 5 – 8 = – 3 M32 = 123−5 = – 5 – 6 = – 11 M33 = 1−134 = 4 + 3= 7 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 = 71−3−19−111−11−17 & |A| = 4 Now, A-1 = 1|A| adj A A-1 = 14 71−3−19−111−11−17 & B = 7−512 Solving X = A-1 B 𝑥𝑦𝑧 = 14 71−3−19−111−11−17 7−512 = 14 7 7+1 −5+(−13)(12)−19 7+(−1) −4+11(12)−11 7+1 −5+7(12) = 14 49−5−36−133+5+132−77+5+84 = 14 8412 𝑥𝑦𝑧 = 213 Hence x = 2 , y = 1, & z = 3

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 provides courses for Mathematics from Class 9 to 12. You can ask questions here.