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The final solution might be more conveniently expressed in terms of complete exponential [[Bell polynomials]] as
:<math> c_{n-k} = \frac{(-1)^{n-k}}{k!} {\mathcal B}_k \Bigl ( \operatorname{tr}A , -1! ~ \operatorname{tr}A^2, 2! ~\operatorname{tr}A^3, \ldots, (-1)^{k-1}(k-1)! ~ \operatorname{tr}A^k\Bigr ) .</math>
==Example==
<math>{\displaystyle A=\left[{\begin{array}{rrr}3&1&5\\3&3&1\\4&6&4\end{array}}\right]}</math>
<math>{\displaystyle {\begin{aligned}B_{0}&=\left[{\begin{array}{rrr}0&0&0\\0&0&0\\0&0&0\end{array}}\right]\qquad \qquad &&&c_{3}&&&&&=&1\\B_{\mathbf {\color {blue}1} }&=\left[{\begin{array}{rrr}1&0&0\\0&1&0\\0&0&1\end{array}}\right]\qquad \qquad &A*B_{1}&=\left[{\begin{array}{rrr}\mathbf {\color {red}3} &1&5\\3&\mathbf {\color {red}3} &1\\4&6&\mathbf {\color {red}4} \end{array}}\right]\qquad \qquad &c_{2}&&&=-{\frac {1}{\mathbf {\color {blue}1} }}\cdot \mathbf {\color {red}10} &&=&-10\\B_{\mathbf {\color {blue}2} }&=\left[{\begin{array}{rrr}-7&1&5\\3&-7&1\\4&6&-6\end{array}}\right]\qquad \qquad &A*B_{2}&=\left[{\begin{array}{rrr}\mathbf {\color {red}2} &26&-14\\-8&\mathbf {\color {red}-12} &12\\6&-14&\mathbf {\color {red}2} \end{array}}\right]\qquad \qquad &c_{1}&&&=-{\frac {1}{\mathbf {\color {blue}2} }}\cdot \mathbf {\color {red}(-8)} &&=&4\\B_{\mathbf {\color {blue}3} }&=\left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right]\qquad \qquad &A*B_{3}&=\left[{\begin{array}{rrr}\mathbf {\color {red}40} &0&0\\0&\mathbf {\color {red}40} &0\\0&0&\mathbf {\color {red}40} \end{array}}\right]\qquad \qquad &c_{0}&&&=-{\frac {1}{\mathbf {\color {blue}3} }}\cdot \mathbf {\color {red}120} &&=&-40\end{aligned}}} </math>
Furthermore <math>{\displaystyle B_{4}=A*B_{3}+c_{0}*I=0}</math> which shows that the calculations where correct.
This implies that the characteristic polynomial of Matrix A is <math>{\displaystyle p_{A}(\lambda )=\lambda ^{3}-10\lambda ^{2}+4\lambda -40.}</math>,the determinant of A is <math>{\displaystyle \det(A)=(-1)^{3}\cdot c_{0}=40}</math> and the inverse of A is
<math>{\displaystyle A^{-1}=-{\frac {1}{c_{0}}}\cdot B_{3}={\frac {1}{40}}\cdot \left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right]=\left[{\begin{array}{rrr}0{,}15&0{,}65&-0{,}35\\-0{,}20&-0{,}20&0{,}30\\0{,}15&-0{,}35&0{,}15\end{array}}\right]} </math>
==An equivalent but distinct expression==
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