# Are there any identities for the determinant of almost upper triangular matrices of the following form?

I've encountered a problem in which I need to compute the determinant of an almost upper triangular matrix of the following form: $$A = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\ 1 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\ 1 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\ 1 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\ 1 & 0 & 0 & 0 & a_{5,5} & \\ \vdots & & & & & \ddots \\ \\ 1 & 0 & 0 & 0 & \dots & 0 & a_{N,N} \end{pmatrix}$$

All matrix entries below the diagonal are zero, except those in the first column, which are equal to one.

The matrix is infinite, so $$N \to \infty$$. I wonder whether there are identities that describe the form of the determinant of this matrix. References to relevant articles are appreciated.

Note that we can write this matrix in the form $$A = B + uv^T$$, where $$B = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\ 0 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\ 0 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\ 0 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\ 0 & 0 & 0 & 0 & a_{5,5} & \\ \vdots & & & & & \ddots \\ \\ 0 & 0 & 0 & 0 & \dots & 0 & a_{N,N} \end{pmatrix}, \quad u = (0,1,\dots,1)^T, \quad v = (1,0,\dots,0)^T.$$ With the matrix determinant lemma, we find that $$\det(A) = \det(B + uv^T) = (1 + v^TB^{-1}u) \det(B) \\ = (1 + v^TB^{-1}u) \cdot a_{22} a_{33} \cdots a_{NN}.$$ From there, it suffices to find $$v^TB^{-1}u$$, i.e. the first entry of $$B^{-1}u$$.
I don't think that there is a nice explicit form for $$v^TB^{-1}u$$, but the answer can be computed very efficiently because the matrix is upper triangular.
• thank you. Maybe the article "On an explicit formula for for inverse triangular matrices" by Baliarsingh and Dutta can help with the determination of the explicit form of $B^{-1}$. Though it seems one needs to compute a lot of determinants with their method, which can get messy Aug 14, 2020 at 20:55