4 a 2 4 3 1 11 s11 s12 s13 s

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Unformatted text preview: 5 Ⱥ Ⱥ 2 12 Ⱥ [A11 ] × [B 11 ] = Ⱥ Ⱥ × Ⱥ Ⱥ = Ⱥ - 2 Ⱥ Ⱥ 5 - 18 Ⱥ 2 6 Ⱥ 6 Ⱥ Ⱥ Ⱥ [A12 ] × [B 21 ] = Ⱥ 4 Ⱥ × [7 1 ] = Ⱥ 28 21 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ 3 Ⱥ Ⱥ 3 Ⱥ Ⱥ 21 - 3 Ⱥ Ⱥ 2 [A 21 ] × [B 11 ] = [1 0 ] × Ⱥ - 3] Ⱥ = [2 6 Ⱥ Ⱥ 5 [A 22 ] × [B 21 ] = [3 ] × [7 1 ] = [21 3 ] [A ] × [B ] Ⱥ 37 = Ⱥ - 6 Ⱥ Ⱥ 2 Ⱥ 12 - 18 -3 Ⱥ Ⱥ 28 Ⱥ + Ⱥ Ⱥ Ⱥ 21 Ⱥ Ⱥ 21 Ⱥ Ⱥ 4 3 3 Ⱥ Ⱥ 65 Ⱥ = Ⱥ Ⱥ Ⱥ 15 Ⱥ Ⱥ 23 Ⱥ Ⱥ 16 Ⱥ - 15 Ⱥ Ⱥ 0 Ⱥ Ⱥ Ⱥ a11 [A]= Ⱥ Ⱥ a 21 a12 Ⱥ Ⱥ; A = a 22 Ⱥ a11 a12 a 21 a 22 ; A = a11a22 - a12 a 21 For example Ⱥ 2 [A] = Ⱥ Ⱥ 1 - 3 Ⱥ Ⱥ ; 4 Ⱥ A = [( 2 )( 4 ) - ( - 3 )(1) ] = 11 Ⱥ S11 S12 S13 Ⱥ [S]= Sij = ȺS21 S22 S23 Ⱥ then Let Ⱥ Ⱥ ȺS31 S32 S33 Ⱥ Ⱥ Ⱥ S11 S = S 21 S31 S12 S 22 S32 S13 S 23 S33 A determinant of order n can be evaluated from the cofactors (Coij) of element Sij from minor Mij of any row I or column j as follows: Ⱥ 3 [ A ] = Ⱥ 1 Ⱥ Ⱥ 9 Ⱥ Co 11 = 0 -1 2 2 Ⱥ Ⱥ 7 Ⱥ ; 6 Ⱥ Ⱥ A= -1 2 7 = ( - 1 ) 2 {( 1 )( 6 ) - ( 7 )( 2 )} = - 20 6 1 + 1` (- 1 ) 1 = (- 1 )1 + 2 ` Co 12 9 Co 13 = 1+ 3` (- 1 ) 1 9 n ∑ a ij Co ij j =1 7 = ( - 1 ) 3 {1 )( 6 ) - ( 7 )( 9 )} = - 57 ( 6 -1 = ( - 1 ) 4 {1 )( 2 ) - ( - 1 )( 9 )} = 11 ( 2 A = ( 3 )( - 20 ) + ( 0 )( - 57 ) + ( 2 )( 11 ) = - 38 [ A Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ = 3 4 1 0 2 1 0 1 A 3 = - 0 - 1 4 - 1 2 - - 1) 0 4 - 1 - 0 2 0 2 3 1 0 4 - 3 1 4 ( Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ 4 3 1 2 3 1 - 0 2 3 3 2 1 1 2 3 1 4 4 - 2 1 0 2 1 0 3 - 1 1 + 1 4 -2 3 2 2 4 3 0 0 -1 1 2 2 1 4 -6 0 0 0 0 0 -2 4 -3 8 3 4 -6 -2 - 12 -2 5 1 2 -3 = 0 0 0 3 -2 -3 16 3 1 = ( 1 )( - 6 )( - 3 )...
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