Ux c is solved by back substitution Here we concentrate on connecting A to U

Ux c is solved by back substitution here we

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Ux = c is solved by back-substitution. Here we concentrate on connecting A to U . The matrices E for step 1, F for step 2, and G for step 3 were introduced in the previous section. They are called elementary matrices , and it is easy to see how they work. To subtract a multiple of equation j from equation i , put the number into the ( i , j ) position . Otherwise keep the identity matrix, with 1s on the diagonal and 0s elsewhere. Then matrix multiplication executes the row operation.
NOT FOR SALE Strang-5060 book April 25, 2005 17:38 33 1.5 Triangular Factors and Row Exchanges 33 The result of all three steps is GFEA = U . Note that E is the first to multiply A , then F , then G . We could multiply GFE together to find the single matrix that takes A to U (and also takes b to c ). It is lower triangular (zeros are omitted): From A to U GFE = 1 1 1 1 1 1 1 1 1 2 1 1 = 1 2 1 1 1 1 . (3) This is good, but the most important question is exactly the opposite: How would we get from U back to A ? How can we undo the steps of Gaussian elimination? To undo step 1 is not hard. Instead of subtracting, we add twice the first row to the second. (Not twice the second row to the first!) The result of doing both the subtraction and the addition is to bring back the identity matrix: Inverse of subtraction is addition 1 0 0 2 1 0 0 0 1 1 0 0 2 1 0 0 0 1 = 1 0 0 0 1 0 0 0 1 . (4) One operation cancels the other. In matrix terms, one matrix is the inverse of the other. If the elementary matrix E has the number in the ( i , j ) position, then its inverse E 1 has + in that position. Thus E 1 E = I , which is equation (4). We can invert each step of elimination, by using E 1 and F 1 and G 1 . I think it’s not bad to see these inverses now, before the next section. The final problem is to undo the whole process at once, and see what matrix takes U back to A . Since step 3 was last in going from A to U, its matrix G must be the first to be inverted in the reverse direction . Inverses come in the opposite order! The second reverse step is F 1 and the last is E 1 : From U back to A E 1 F 1 G 1 U = A is LU = A . (5) You can substitute GFEA for U , to see how the inverses knock out the original steps. Now we recognize the matrix L that takes U back to A . It is called L , because it is lower triangular . And it has a special property that can be seen only by multiplying the three inverse matrices in the right order: E 1 F 1 G 1 = 1 2 1 1 1 1 1 1 1 1 1 1 = 1 2 1 1 1 1 = L . (6) The special thing is that the entries below the diagonal are the multipliers = 2 , 1, and 1. When matrices are multiplied, there is usually no direct way to read off the answer. Here the matrices come in just the right order so that their product can be written down immediately. If the computer stores each multiplier i j —the number that multiplies the pivot row j when it is subtracted from row i

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