Determinant of symmetric matrix Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?  
\begin{pmatrix}
  2 & 1 &1 &1&1 \\
  1 & 2 & 1& 1 &1\\
  1& 1 & 2 & 1  &1\\
 1&1 &1 &2&1\\
 1&1&1&1&-2
 \end{pmatrix}
 A: Look at the matrix $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -3 \end{pmatrix}.$$ It has rank two and its nonzero eigenvalues have sum $1$ (the trace) and product $$\det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} = -16$$
So the characteristic polynomial is $t^3 (t^2 - t - 16)$. Evaluate this at $t = -1$ and you get $$\det(-I - A) = (-1)^5 \det \begin{pmatrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & -2 \end{pmatrix}.$$
A: You can substract the first row from every other rows and get matrix of form:
$$\begin{pmatrix}
  2 & 1 &1 &1&1 \\
  -1 & 1 & 0& 0 &0\\
  -1& 0 & 1 & 0  &0\\
 -1&0 &0 &1&0\\
 -1&0&0&0&-3
 \end{pmatrix}.$$
Computing the determinant is now much easier.
A: Call your matrix $A$ and let $J_n$ denotes the all-one matrix of size $n$. By Laplace expansion along the last row, we have
\begin{align*}
\det(A)
&=\left(\sum_{j=1}^\color{red}{4}(-1)^{5+j}a_{5j}M_{5j} + 2M_{55}\right) - 4M_{55}\\
&=\det(I_5+J_5)-4\det(I_4+J_4).
\end{align*}
As $\det(aI_n+bJ_n)=a^{n-1}(a+nb)$, we get $\det(A)=6-4(5)=-14$.
