In exercise [1, p. 34 exercise 2.1] we are asked to prove that the inverse of a lower triangular matrix is also lower triangular by using relation [1, p. 23, (2.30)].
Solution: The determinant of a lower triangular matrix is given by [1, p. 20, (2.16)]
det(\mathbf{L})=\prod_{i=1}^{N}{l_{ii}} (1)

, as also by [1, p. 18, (2.1)]
det(\mathbf{L})=\sum\limits_{j=1}^{N}l_{kj} {c}_{kj}. (2)

In the previous equation k is any positive integer smaller or equal than the size N of the matrix. The element c_{kj} is given by c_{kj}=(-1)^{(j+k)} det(\mathbf{M}_{kj}), where \mathbf{M}_{kj} is the minor matrix obtained by removing the k^{th} row and j^{th} column from the matrix \mathbf{L}. Now the elements \{l_{kj}\} of the lower triangular matrix \mathbf{L} are zero for j > k. Equation (2) can be rewritten for the lower triangular matrix as:
det(\mathbf{L})=\sum\limits_{j=1}^{k}l_{kj} c_{kj}. (3)

Using (1) and (3) we derive the following relations:
\prod_{i=1}^{N}{l_{ii}}=\sum\limits_{j=1}^{k} l_{kj} {c}_{kj} (4)
l_{kk} (\prod_{i=1,i\neq k}^{N}{l_{ii}}-c_{kk})=\sum\limits_{j=1}^{k-1}l_{kj} c_{kj} (5)

The left hand side of the previous equation equals zero because c_{kk}=\prod_{i=1,i\neq k}^{N}{l_{ii}} which can be derived by utilizing the fact that c_{kk} is the determinant of the lower triangular matrix that is obtained by \mathbf{L} when the k^{th} row and column are removed. Thus the following conditions must be satisfied, for all k:
\sum\limits_{j=1}^{k-1}l_{kj} {c}_{kj}=0 (6)

We note that the c_{kj}, j=1,...(k-1) are independet of the l_{kj}, j=1,...(k-1), because c_{kj} is obtained by \mathbf{L} when the k^{th} row is removed where the l_{kj} are located. It follows that the l_{kj}, j=1,...(k-1) do not contribute to the result of c_{kj}. But because the l_{kj}, k=1,...,(k-1) are not restricted to any values we derive the neccesity that
c_{kj}=0, j=1,...(k-1). (7)

The matrix \mathbf{C}=[c_{ij}] is an upper triangular matrix as c_{kj}=0 for k > j. The inverse of the lower triangular matrix is given by [1, p. 23, (2.30)]: \mathbf{L}^{-1}=\frac{\mathbf{C}^T}{det(\mathbf{L})} which is also a lower triangular matrix because \mathbf{C}^T is lower triangular. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.