Ex 4.5, 9 - Find inverse of [2 1 3 4 -1 0 -7 2 1] - Class 12 - Ex 4.5

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.5, 9 Find the inverse of each of the matrices (if it exists). 2 1 3 4 1 0 7 2 1 Let A = 2 1 3 4 1 0 7 2 1 We know that A 1 = 1 |A| (adj A) exists if |A| 0 Step 1: Calculate |A| |A| = 2 1 3 4 1 0 7 2 1 = 2 1 0 2 1 1 4 0 7 1 + 3 4 1 7 2 = 2( 1 0) 1 (4 0) + 3 (8 7) = 2 ( 1) 1 (4) + 3 (1) = 3 Since 0 , A 1 exists Step 2: Calculating adj (A) adj (A) = A11 A21 A31 A12 A22 A32 A33 A23 A33 A = 2 1 3 4 1 0 7 2 1 M11 = 1 0 2 1 = 1(1) 0(2) = -1 M12 = 4 0 7 1 = 4(1) 0( 7) = 4 M13 = 4 1 7 2 = 4(2) (-7)(-1) = 1 M21 = 1 3 2 1 = 1(1) 2(3) = 5 A11 = ( 1)1 + 1 M11 = ( 1)2 ( 1) = 1 ( 1) = 1 A12 = ( 1)1+2 M12 = ( 1)3 ( 4) = ( 1) (4) = 4 A13 = ( 1)1+3 M13 = ( 1)4 (1) = 1 A21 = ( 1)2+1 M21 = ( 1)3 ( 5) = ( 1)( 5) = 5 A22 = ( 1)2+2 M22 = ( 1)4 ( 23) = 23 A23 = ( 1)2+3 M23 = ( 1)5 11 = 1 (11) = 11 A31 = ( 1)3+1 M31 = ( 1)4 (4) = 1 (3) = 3 A32 = ( 1)3+2 M32 = ( 1)5 ( 12) = ( 1) ( 12) = 12 A33 = ( 1)3+3 M33 = ( 1)6 ( 6) = 1 . ( 6) = 6 Thus, adj (A) = A11 A21 A31 A12 A22 A32 A33 A23 A33 = 1 5 3 4 23 12 1 11 6 Step 3: Calculate inverse A 1 = 1 |A| ( adj (A)) = 1 3 1 5 3 4 23 12 1 11 6 =

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