<s>[INST] How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
 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{pmatrix}
I expanded it out and got $349$ but I feel there must be some trick to easily compute it. 
 [/INST] There is a general pattern for this kind of matrices. Define 
$$
A_n:=\mathbf 1_n+\operatorname{diag}(1,2,\dots,n)=\pmatrix{2&1&1&\dots &1&1\\
1&3&1&\dots&1&1\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&1\\
1&1&1&\dots &1&n+1},
$$
where $\mathbf 1_n$ is the $n\times n$ matrix whose entries are one.
View the last column as $\pmatrix{1\\1\\\vdots\\1\\1}+n\pmatrix{0\\0\\\vdots\\0\\1}$. 
Denoting by $D_n$ the determinant of $A_n$, we get 
$$
D_n=\det \pmatrix{2&1&1&\dots &1&1\\
1&3&1&\dots&1&1\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&1\\
1&1&1&\dots &1&1}+n\det\pmatrix{2&1&1&\dots &1&0\\
1&3&1&\dots&1&0\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&0\\
1&1&1&\dots &1&1}.
$$
The first determinant can be computed in the following way: subtract to  each column the last one to be reduced to compute the determinant of an upper diagonal matrix. The second determinant is $D_{n-1}$ so we end up with the recurrence relation 
$$
D_n=nD_{n-1}+(n-1)!.
$$
This can be solved by letting $a_n:=D_n/n!$: we get $a_n=a_{n-1}+1/n$ hence 
$$D_n=\left(1+\sum_{j=1}^n\frac 1j\right)n!.$$
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