[–] 3 points4 points  (1 child)

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That looks fine. you took two arbitrary elements of L2 and showed that the product is also in L2. For part (2), you could just use the formula for the inverse of a 2x2. If A is given by [{a, b};{c, d}], then the inverse is 1/det(A) [{d, -b};{-c, a}], i.e. switch the main diagonal elements, take the opposite of the off-diagonal elements, divide everything by the determinant. This, of course, assumes that the inverse exists(in this case, that a and d are nonzero)

[–] 1 point2 points  (0 children)

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Actually, there is a mistake in your work for part (1). The approach is ok, but you have to make sure to multiply the matrices properly. Check the term in the second row, first column of the product

[–] 2 points3 points  (0 children)

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|a 0| |d 0| = | ad       0  |
|b c| |e f|   | bd + ce  cf |

[–] 0 points1 point  (0 children)

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Another approach to seeing it: notice that solving a lower triangular system of equations works by forward substitution. The first equation (first row of the matrix) gives you the first variable, then with that known the second equation (second row) gives you the second variable, etc.

That makes the first variable of the solution a function of the constant of the first equation, then the second variable a function of the constants of the first two equations, etc. Which is what the inverse being lower triangular means.
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