# Ex 4.6, 12 - 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, 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 1 1 2 1 3 1 1 1 = 4 0 2 Hence A = 1 1 1 2 1 3 1 1 1 , X = & B = 4 0 2 Step 2 Calculate |A| |A| = 1 1 1 2 1 3 1 1 1 = 1 1 3 1 1 ( 1) 2 3 1 1 + 1 2 1 1 1 = ( 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 = A11 A12 A13 A21 A22 A23 A31 A32 A33 = A11 A21 A31 A12 A22 A32 A13 A23 A33 A = 1 1 1 2 1 3 1 1 1 M11 = 1 3 1 1 = 1 + 3 = 4 M12 = 2 3 1 1 = 2 + 3 = 5 M13 = 2 1 1 1 = 2 1= 1 M21 = 1 1 1 1 = 1 1 = 2 M22 = 1 1 1 1 = 1 1 = 0 M23 = 1 1 1 1 = 1 + 1 = 2 M31 = 1 1 1 3 = 3 1 = 2 M32 = 1 1 2 3 = 3 4 = 5 M33 = 1 1 2 1 = 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 = 1 2+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 = 4 2 2 5 0 5 1 2 3 & |A| = 10 So, A-1 = 1 |A| adj A A-1 = 1 10 4 2 2 5 0 5 1 2 3 & B = 4 0 2 Now, solving X = A-1 B = 1 10 4 2 2 5 0 5 1 1 3 4 0 2 = 1 10 4 4 +2 0 +2(2) 0 4 +0 0 +5(2) 2 4 +1 0 +3(2) = 1 10 16+0+4 20+0+10 4+0+6 = 1 10 20 10 10 = 2 1 1 Hence, x = 2 , y = 1, & z = 1

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

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.