\[A= \begin{bmatrix} 1 &2 & 1 & 0\\ 1& 0 & 1 & 0\\ 2& 1& 1 & 0\\ 3& 1& 0 & 1 \end{bmatrix}\]
\[A= \begin{bmatrix} 1 &2 & 1 & 0\\ 1& 0 & 1 & 0\\ 2& 1& 1 & 0\\ 3& 1& 0 & 1 \end{bmatrix} ; I =\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 & 0\\ 0& 0 & 0 & 1 \end{bmatrix}\]
pA(λ) = det(A???λI4)
\[\mathbf{pA({\lambda})} = \mathbf{det(\left[\begin{matrix} 1 & 2 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1\end{matrix}\right] + (-{\lambda} \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{matrix}\right]))}\]
\[\mathbf{pA({\lambda})} = \mathbf{det(\left[\begin{matrix} 1-\lambda & 2 & 1 & 0 \\ 1 & 0-\lambda & 1 & 0 \\ 2 & 1 & 1-\lambda & 0 \\ 3 & 1 & 0 & 1-\lambda \end{matrix}\right] }\]
\[=(1-\lambda )\begin{vmatrix} 0-\lambda & 1 & 0\\ 1& 1-\lambda & 0\\ 1& 0& 1-\lambda \end{vmatrix}+2\begin{vmatrix} 1 & 1 & 0\\ 2 & 1-\lambda & 0\\ 3 & 0 & 1-\lambda \end{vmatrix}+1\begin{vmatrix} 1 & 0-\lambda & 0\\ 2 & 1 & 0\\ 3 & 1 & 1-\lambda \end{vmatrix}\]
Solution by r:
library(pracma)## Warning: package 'pracma' was built under R version 3.3.3A <- matrix(c(1,1,2,3,2,0,1,1,1,1,1,0,0,0,0,1), nrow = 4, ncol = 4, byrow = TRUE)
charpoly(A)## [1] 1 -3 -2 2 2