    1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Ex 4.5

Transcript

Ex 4.5, 7 Find the inverse of each of the matrices (if it exists). 1 2 3 0 2 4 0 0 5 Let A = 1 2 3 0 2 4 0 0 5 We know that A 1 = 1 |A| (adj A) exists if |A| 0 Step 1: Calculate |A| |A| = 1 2 3 0 2 4 0 0 5 = 1 2 4 0 5 2 0 4 0 5 3 0 2 0 0 = 1(10 0) 2 (0 0) 3 (0 0) = 1 (10) 0 0 = 10 Since |A| 0 A 1 exists Step 2: Calculating adj A adj A = = A11 A21 A31 A12 A22 A32 A33 A23 A33 A = 1 2 3 0 5 4 0 0 5 M11 = 2 4 0 5 = 2(5) 0(4) = 10 M12 = 0 4 0 5 = 0(5) 0(4) = 0 M13 = 0 2 0 0 = 0 0 = 0 M21 = 2 3 0 5 = 10 0 = 10 A11 = ( 1)1-1 M11 = ( 1)2 (10) = 1(10) = 10 A12 = ( 1)1+2 M12 = ( 1)3 0 = 0 A13 = ( 1)1+3 M13 = ( 1)4 0 = 0 A21 = ( 1)2+1 M21 = ( 1)3 (10) = 1 (10) = 10 A22 = ( 1)2+2 M22 = ( 1)4 5 = 5 A23 = ( 1)2+3 M23 = ( 1)4 0 = 0 A31 = ( 1)3+1 M31 = ( 1)4 2 = 2 A32 = ( 1)3+2 M32 = ( 1)5 4= 1 (4) = 4 A33 = ( 1)3+3 M33 = ( 1)6 2 = 2 Thus adj (A) = A11 A21 A31 A12 A22 A32 A33 A23 A33 = 10 10 2 0 5 4 0 0 2 Step 3: Calculate inverse Now , A 1 = 1 |A| ( adj (A)) =

Ex 4.5 