# Ex 4.6, 5 - Chapter 4 Class 12 Determinants

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.6, 5 Examine the consistency of the system of equations. 3x − y − 2z = 2 2y − z = −1 3x − 5y = 3 The system of equations can be written as 3x − y − 2z = 2 0x + 2y − z = −1 3x − 5y + 0z = 3 Step1 Write equation as AX = B 3−1−202−13−50 𝑥𝑦𝑧 = 213 Hence A = 3−1−202−13−50, X = 𝑥𝑦𝑧 & B = 213 Step 2 Calculate |A| |A| = 3−1−202−13−50 = 3 2−1−50 – ( – 1) 0−132 + 2 023−5 = 3(0 – 5) + 1 (0 + 3) – 2 (0 – 6) = 3 ( – 5) + 1 (3) – 2 ( – 6) = – 15 + 3 + 12 = – 15 + 15 = 0 Since |A| = 0, We calculate adj A (B) adj A = A11 A12 A13 A21 A22 A23 A31 A32 A33′= A11 A21 A31 A12 A22 A32 A13 A32 A33 A = 3−1−202−13−50 M11 = 2−1−50 = 0 – (5) = – 5 M12 = 0−130 = 0 + 3 = –3 M13 = 623−5 = 0 – 6 = – 6 M21 = −1−2−50 = 0 – 10 = – 10 M22 = 3−230 = 0 + 6 = 6 M23 = 3−13−5 = – 15 +3 = – 12 M31 = −1−22−1 = 1 + 4 = 5 M32 = 3−20−1 = – 3 + 0 = – 3 M33 = 3−102 = 6 + 0 = 6 A11 = ( – 1)1+1 . M11 = ( – 1)2 . ( – 5) = – 5 A12 = ( – 1)1+2 . M12 = ( – 1)3 . 3 = – 3 A13 = ( – 1)1+3 . M13 = ( – 1)4 . ( – 6) = – 6 A21 = ( – 1)2+1 . M21 = ( – 1)3 . ( – 10) = 10 A22 = ( – 1)2+2 . M22 = ( – 1)4 . 6 = 6 A23 = ( – 1)2+3 . M23 = ( – 1)5 . ( – 12) = 12 A31 = ( – 1)3+1 . M31 = ( – 1)4 . 5 = 5 A32 = ( – 1)3+2 . M32 = ( – 1)5 . ( – 3) = 3 A33 = ( – 1)3+3 . M33 = ( – 1)6 . (6) = 6 Thus, adj (A) = A11 A21 A31 A12 A22 A32 A13 A32 A33 = 3−1−202−13−50 Now, adj (A) . B Putting values = −5105−363−6126 2−13 = −5 2+10 −1+5 3−3 2+6 −1+3 3−6 2+12 −1+6 3 = −10−10+15−6−6+9−12−12+18 = −5−3−6 Thus, adj A . B ≠ O Therefore, |A| = 0 & (adj A) B ≠ O, Thus, the given system equation is inconsistent & the system of equations has no solution

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.