0 1 0 0 1 1 0 2 0 5 1 0 0 1 0 1 divide row 3

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Unformatted text preview: ( - 8 ) = 144 3 -2 4 -7 -6 4 -2 0 4 -6 0 0 4 -3 0 0 0 = 3 -2 -3 -8 = Ⱥ1 -1 2 Ⱥ [A ] = Ⱥ3 0 1 Ⱥ; Ⱥ Ⱥ Ⱥ1 0 2 Ⱥ Ⱥ Ⱥ Ⱥ1 -1 2 1 0 Ⱥ Ⱥ3 0 1 0 1 Ⱥ1 0 2 0 0 Ⱥ Augment matrix A with the identity matrix 0 Ⱥ Ⱥ Ⱥ → Multiply row 1 by 3 and 0 Ⱥ Ⱥ subtract it from row 2 and subtract row 1 from row 1 Ⱥ Ⱥ 3 Ⱥ1 -1 2Ⱥ [A]= Ⱥ3 0 1Ⱥ Ⱥ → Ⱥ Ⱥ Ⱥ Ⱥ1 0 2Ⱥ Ⱥ Ⱥ Augment the matrix with the identity matrix Ⱥ1 -1 2 1 0 0Ⱥ Multiply row 1 by 3 and Ⱥ Ⱥ Ⱥ → subtract it from row 2 and 3 0 1 0 1 0Ⱥ Ⱥ Ⱥ subtract row 1 from row 3 Ⱥ1 0 2 0 0 1Ⱥ Ⱥ Ⱥ Ⱥ -1 2 1 0 0Ⱥ 1 Ⱥ ȺȺ→ Interchange rows 3 and 2 -Ⱥ Ⱥ0 3 5 3 1 0Ⱥ Ⱥ0 1 0 -1 0 1Ⱥ Ⱥ Ⱥ Ⱥ 1 Ⱥ Ⱥ0 Ⱥ0 Ⱥ -1 2 1 0 0Ⱥ -1 0 1ȺȺ→ Multiply row 1 by 3 and add Ⱥ 10 Ⱥ it to row 3 3 -5 -3 1 0Ⱥ Ⱥ Ⱥ 1 Ⱥ 0 Ⱥ Ⱥ 0 Ⱥ -1 1 0 2 0 -5 1 0 0 Ⱥ -1 0 1 ȺȺ→ Divide row 3 by 5 Ⱥ Ⱥ 0 1 -3Ⱥ Ⱥ Ⱥ 1 Ⱥ -1 2 1 Ⱥ 1 0 -1 0 Ⱥ 0 Ⱥ 0 1 0 Ⱥ Ⱥ 0 0 Ⱥ Add row 2 to row 1 and Ⱥ 0 1 ȺȺ→ multiply row 3 by 2 and 1 3Ⱥ subtract it from row 1 - - Ⱥ 5 5Ⱥ Ⱥ Ⱥ1 0 0 Ⱥ0 1 0 Ⱥ Ⱥ0 0 1 Ⱥ Ⱥ Ⱥ Ⱥ 0 Ⱥ -1 Ⱥ Ⱥ 0 Ⱥ Ⱥ 2 5 0 1 5 0 -1 0 2 5 0 1 5 1 Ⱥ - Ⱥ 5 Ⱥ Ⱥ 1 Ⱥ Ⱥ → 3 - Ⱥ 5 Ⱥ Ⱥ 1 Ⱥ - Ⱥ 5 Ⱥ 1 Ⱥ = [A]-1 3 - Ⱥ 5 Ⱥ Ⱥ The 3 by 3 matrix on the right hand side is the inverse [A]-1 of matrix [A] [S ]-1 = [ Co ]T S where [Co] is a matrix...
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This note was uploaded on 02/05/2014 for the course CE 221 taught by Professor Buch during the Fall '08 term at Michigan State University.

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