<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] 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.
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