Problem 2.6.8. This problem shows how the elimination matrix inverses E¹ multiply to give L. You see this best when A is already lower triangular with 1's on the diagonal. Then U = I. 1 0 0 a 1 0 b с 1 The elimination matrices E21, E31, E32 contain –a then −b then -c. A = L = (a) Multiply E32 E31 E21 to find the single matrix E that produces EA = I. (b) Multiply E¹E1¹ E32¹ to bring back L. The multipliers a, b, c are mixed up in E but perfect in L!

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