       1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Ex 4.6

Transcript

Step 2 Calculate |A| |A| = 2 1 1 1 2 1 0 3 5 = 2 2 1 3 5 1 1 1 0 5 + 1 1 2 0 3 = 2 (10 + 3 ) 1 ( 5 + 0) + 1 (3 0) = 2 (13) 1 ( 5 ) + 21 (3) = 34 Thus, |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 A-1 = 1 |A| adj (A) adj (A) = A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 = A 11 A 21 A 31 A 12 A 22 A 23 A 13 A 32 A 33 A = 2 1 1 1 2 1 0 3 5 M11 = 2 1 3 5 = 15 + 3 = 13 M12 = 1 1 0 5 = 5 + 0 = 5 M13 = 1 2 0 3 = 3 + 0 = 3 M21 = 1 1 3 5 = 5 3 = 8 M22 = 2 1 0 5 = 10 + 0 = 10 M23 = 2 1 0 3 = 6 + 0 = 6 M31 = 1 1 2 1 = 1 + 2 = 1 M32 = 2 1 1 1 = 2 1 = 3 M33 = 2 1 1 2 = 4 1 = 5 A11 = ( 1)1+1 . M11 = ( 1 )2 . (13) = 13 A12 = ( 1)1+2 . M12 = ( 1 )3 . ( 5) = 5 A13 = ( 1)1+3 . M13 = ( 1 )4 . (3) = 3 A21 = ( 1)2+1 . M21 = ( 1 )3 . ( 8) = 8 A22 = ( 1)2+2 . M22 = ( 1 )4 . ( 10) = 10 A23 = ( 1)2+3 . M23 = ( 1 )5 . (6) = 6 A31 = ( 1)3+1 . M31 = ( 1 )4 . (1) = 1 A32 = ( 1)3+2 . M32 = ( 1 )5 . ( 3) = 3 A33 = ( 1)3+3 . M33 = ( 1 )6 . ( 5) = 5 Thus, adj (A) = 13 8 1 5 10 3 3 6 5 Now, A-1 = 1 |A| adj A Putting values = 1 34 13 8 1 5 10 3 3 6 5 Also, X = A-1 B Putting values = 1 34 13 8 1 5 10 3 3 6 5 1 3 2 9 = 1 34 13 1 +8 3 2 +1 9 5 1 + 10 3 2 +3 9 3 1 + 6 3 2 + 5 9 = 1 34 13+12+9 5 15+27 3 9 45 = 1 34 34 17 51 = 1 1 2 3 2 Hence x = 1, y = & z =

Ex 4.6 