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**Deveno** Actually, this problem is *not* that hard, provided it is two separate problems. The second one, for example, is to find U^{-1}, where:

$\displaystyle U = \begin{bmatrix}1&a&b\\0&1&c\\0&0&1 \end{bmatrix}$.

you should know (hopefully) that the product of two upper-triangular matrices with diagonal entries of 1, is another such matrix. since I (the identity matrix) is also upper triangular, with diagonal entries all 1, it seems reasonable to conjecture that U^{-1} is also upper-triangular with 1's on the diagonal. so we want to solve THIS equation:

$\displaystyle \begin{bmatrix}1&a&b\\0&1&c\\0&0&1 \end{bmatrix} \begin{bmatrix}1&x&y\\0&1&z\\0&0&1 \end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

for x,y and z, in terms of a,b and c. you can do this.