Matrix of Ones with Diagonal of Integers

Question

My teacher posed a question to the class today asking us to find the determinant of the following matrix...

\begin{bmatrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 & 1 \\ 1 & 1 & 4 & 1 & 1 \\ 1& 1 & 1 & 5 & 1 \\ 1&1&1&1&6 \end{bmatrix}

using a simple trick that doesn't involve transforming it into reduced row echelon form. For the life of me I've been unable to figure it out. Does anyone know the trick, or even which steps I should take to make my teachers supposed method more apparent?

Well, OEIS tells me that the determinant of an $n\times n$ matrix of this form is $$n!\left(1+\sum\limits_{i=1}^n\frac{1}{i}\right)$$ but I haven't found a nice tricky way to see it. — Peter Woolfitt, Mar 17 '15 at 20:43
This is a particular case of http://math.stackexchange.com/questions/2110766/calculating-determinant-with-different-numbers-on-diagonal-and-x-everywhere-else . — darij grinberg, Jan 24 '17 at 22:46

score 3 · Answer 1 · answered Mar 17 '15 at 21:21

Call your structured matrix $M_n$. Use the matrix determinant lemma Matrix-Determinant Lemma: \begin{align} \text{det}(A+uv^T) = (1+v^\mathrm{T}A^{-1}u)\text{det}(A) \end{align} and notice that the matrix can be decomposed into $M_n = \text{diag}(1,2,3,4,...,n) + ee^T$, where $e\in \mathbb{R}^n$ is a vector full of ones, i.e $e_j = 1,\forall j$. Apply the theorem and you get: \begin{align} \text{det}(M_n) = (1+\sum\limits^{n}_{k=1} \frac{1}{k})n! \end{align}

Matrix of Ones with Diagonal of Integers

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