Let $$ P = \begin{pmatrix} 0 & 1 & 1 & 1 & 1 \ -1 & 0 & 1 & 1 & 1 \ -1 & -1 & 0 & 1 & 1 \ -1 & -1 & -1 & 0 & 1 \ -1 & -1 & -1 & -1 & 0 \end{pmatrix} $$ If $ \lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5 $ are eigenvalues of $ P $, then $ \prod_{i=1}^{5} \lambda_i = $ __________ (answer in integer).

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Step 1: Recognize the structure of the matrix $ P $. The matrix $ P $ is a $ 5 \times 5 $ matrix where each off-diagonal element is $ -1 $, and diagonal elements are $ 0 $. This matrix represents the Laplacian matrix of the complete graph $ K_5 $, where each node is connected to every other node. The Laplacian matrix for a complete graph $ K_n $ has the following properties: One eigenvalue is $ n - 1 $, where $ n $ is the number of vertices (in this case, $ n = 5 $, so one eigenvalue is 4). The remaining eigenvalues are $ -1 $, with multiplicity $ n - 1 $. Thus, the eigenvalues of $ P $ are: $$ \lambda_1 = 4, \quad \lambda_2 = \lambda_3 = \lambda_4 = \lambda_5 = -1. $$ Step 2: Compute the product of the eigenvalues.   The product of the eigenvalues is: $$ \prod_{i=1}^{5} \lambda_i = 4 \times (-1)^4 = 4 \times 1 = 4. $$ However, there is a critical point we missed earlier: The matrix $ P $ as it is given actually represents a signed Laplacian matrix for a graph with negative weights. This would imply that the eigenvalues are not simply $ 4 $ and $ -1 $. In fact, the product of eigenvalues will be $ 0 $ because one eigenvalue will be $ 0 $, as this is a property of signed Laplacians where there is always at least one eigenvalue equal to $ 0 $ due to the row-sum property of Laplacian matrices.  Step 3: Correct the product of eigenvalues.   Since one of the eigenvalues is $ 0 $, the product of the eigenvalues is: $$ \prod_{i=1}^{5} \lambda_i = 0. $$  Thus, the correct answer is $ \boxed{0} $.

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