Determinant of upper triangular matrix with first column not $0$

Question

I have a Matrix $A \in \mathbb{R}^{n \times n}$ which is of the form $A$ = \begin{pmatrix} a_{11} & a_{12} & a_{13} & ... & a_{1n}\\ a_{21} & a_{22} & a_{23} & ... & a_{2n}\\ a_{31} & 0 & a_{33} & ... & a_{3n}\\ ... & 0 & 0 & ... & a_{4n}\\ ... & ... & ... & ... & ... \\ a_{n1} & 0 & ... & 0 & a_{nn}\\ \end{pmatrix}

So it is upper triangular except for the elements $a_{i1}$ which are nonzero. Is there an easy formula for its determinant? I tried to do laplace expansion along the first column but it seemed more difficult then expected.

Answer 1

If you move the first column to the right, the determinant remains the same (up to a sign) and the matrix becomes an upper Hessenberg form. Determinants of matrices of this kind are usually found by iterative methods. I don't think there is any evaluation formula as easy as the one for triangular matrices.