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. 
 A: Use the following rules:


*

*Adding a multiple of one row to another row does not change the determinant.

*If $B$ is obtained by multiplying a row of $A$ by a constant $c$ then $\det B=c \det A$


Start by subtracting the first row from each of the other rows, we get  $$A=\begin{pmatrix}
 2 & 1 & 1  & 1 & 1\\ 
 -1 & 2 & 0 & 0 & 0\\ 
 -1&  0 &  3 &  0 & 0\\ 
 -1 &  0 &  0 &  4 & 0\\ 
 -1 &  0 &  0 &  0 & 5
\end{pmatrix}$$ which has the same determinant as your matrix. Even just doing this makes the determinant much easier to calculate but we can go further. Divide the second, third, fourth and fifth rows by their corresponding diagonal term, we get $$B=\begin{pmatrix}
 2 & 1 & 1  & 1 & 1\\ 
 -1/2 & 1 & 0 & 0 & 0\\ 
 -1/3&  0 &  1 &  0 & 0\\ 
 -1/4 &  0 &  0 &  1 & 0\\ 
 -1/5 &  0 &  0 &  0 & 1
\end{pmatrix}$$ which by rule #$2$, has determinant $\det B =\frac 1{120}\det A$. Finally, subtract each of the other rows from the first row, we get 
$$B'=\begin{pmatrix}
 147/60 & 0 & 0  & 0 & 0\\ 
 -1/2 & 1 & 0 & 0 & 0\\ 
 -1/3&  0 &  1 &  0 & 0\\ 
 -1/4 &  0 &  0 &  1 & 0\\ 
 -1/5 &  0 &  0 &  0 & 1
\end{pmatrix}$$ where $\det B'=\det B$ by rule #$1$. This is a lower triangular matrix so the determinant is simply the product of the diagonal elements, i.e. $\det B' = 147/60$. Therefore $$\det A = 120 \det B=120 \det B' =120 \cdot \frac{147}{60}=394.$$ This is generally the way to go if you're calculating determinants of large matrices by hand, just be sure to keep track of each time you multiply a row by something so you can get back to the original determinant. Also, the symmetric property makes row-reduction easier but you can do this procedure for any matrix.
A: Your matrix is in a very nice form. As an alternate answer, you can use the matrix determinant lemma in this case to do the calculation quite cleanly. 
Let $\mathbf{M}$ be the matrix you provided. Then $\mathbf{M}$ = $\mathbf{A} + \mathbf{j}\mathbf{j^t}$,
where
$$
\mathbf{A} =
\begin{pmatrix}
 1 &  0 &  0 &  0 & 0\\ 
 0 &  2 &  0 &  0 & 0\\ 
 0 &  0 &  3 &  0 & 0\\ 
 0 &  0 &  0 &  4 & 0\\ 
 0 &  0 &  0 &  0 & 5
\end{pmatrix}
$$ and
$$
\mathbf{j} =
\begin{pmatrix}
  1\\ 
 1\\ 
 1\\ 
 1\\ 
 1
\end{pmatrix}.
$$
From applying the lemma, we have that $\text{det}(\mathbf{M}) = (1 + \mathbf{j^tA^{-1}j}) \ \text{det}(\mathbf{A})$.
Since $\mathbf{A}$ is diagonal in this case, the determinant is very easy to compute. It's just $5*4*3*2*1=120$. Moreover, 
$$
\mathbf{A^{-1}} =
\begin{pmatrix}
 1 &  0 &  0 &  0 & 0\\ 
 0 &  \frac{1}{2} &  0 &  0 & 0\\ 
 0 &  0 &  \frac{1}{3} &  0 & 0\\ 
 0 &  0 &  0 &  \frac{1}{4} & 0\\ 
 0 &  0 &  0 &  0 & \frac{1}{5}
\end{pmatrix}
$$ 
and since $\mathbf{j}$ is just a vector of ones, it's easy to see that 
$$\mathbf{j^tA^{-1}j} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$$.
Thus, we see that
$$ \text{det}(\mathbf{M}) = (1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}) * 120 = 394$$ 
after applying some simplification or a calculator.
A: 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 
$$D_n=\left(1+\sum_{j=1}^n\frac 1j\right)n!.$$
