# Ex 4.6, 14 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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

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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.