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Using elementary transformations, find the inverse of the matrix`[[2,-3, 3],[ 2, 2, 3],[ 3,-2, 2]]` |
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Answer» Let `A = [[2,-3,3],[2,2,3],[3,-2,2]]` We know, `A = AI` `:. [[2,-3,3],[2,2,3],[3,-2,2]] = A[[1,0,0],[0,1,0],[0,0,1]]` Applying `C_2->C_2+C_3` on both sides, `[[2,0,3],[2,5,3],[3,0,2]] = A[[1,0,0],[0,1,0],[0,1,1]]` Applying `C_1->C_3-C_1` on both sides, `[[1,0,3],[1,5,3],[-1,0,2]] = A[[-1,0,0],[0,1,0],[1,1,1]]` Applying `C_3->C_3-3C_1` on both sides, `[[1,0,0],[1,5,0],[-1,0,5]] = A[[-1,0,3],[0,1,0],[1,1,-2]]` Applying `C_3->C_2/5 and C_3->C_3/5` on both sides, `[[1,0,0],[1,1,0],[-1,0,1]] = A[[-1,0,3/5],[0,1/5,0],[1,1/5,-2/5]]` Applying `C_1->C_1 - C_2` on both sides, `[[1,0,0],[0,1,0],[-1,0,1]] = A[[-1,0,3/5],[-1/5,1/5,0],[4/5,1/5,-2/5]]` Applying `C_1->C_1 + C_3` on both sides, `[[1,0,0],[0,1,0],[0,0,1]] = A[[-2/5,0,3/5],[-1/5,1/5,0],[2/5,1/5,-2/5]]` This is the form, `I = A A^-1` `:. A^-1 = [[-2/5,0,3/5],[-1/5,1/5,0],[2/5,1/5,-2/5]]` |
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