Faddeev–LeVerrier algorithm: Difference between revisions

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Example: copyedit. * for matrix multiplication is Highly deprecated. Compressed format, displaying pathologically across several platforms
Example: aligned with above notation... M instead of B!
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==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_M_{0}&=\left[{\begin{array}{rrr}0&0&0\\0&0&0\\0&0&0\end{array}}\right]\quad &&&c_{3}&&&&&=&1\\B_M_{\mathbf {\color {blue}1} }&=\left[{\begin{array}{rrr}1&0&0\\0&1&0\\0&0&1\end{array}}\right] \quad &A~B_M_{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 &c_{2}&&&=-{\frac {1}{\mathbf {\color {blue}1} }} \mathbf {\color {red}10} &&=&-10\\B_M_{\mathbf {\color {blue}2} }&=\left[{\begin{array}{rrr}-7&1&5\\3&-7&1\\4&6&-6\end{array}}\right]\qquad &A~B_M_{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 &c_{1}&&&=-{\frac {1}{\mathbf {\color {blue}2} }} \mathbf {\color {red}(-8)} &&=&4\\B_M_{\mathbf {\color {blue}3} }&=\left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right]\qquad \qquad &A~B_M_{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} }}\mathbf {\color {red}120} &&=&-40\end{aligned}}} </math>
 
Furthermore <math>{\displaystyle B_M_{4}=A~B_M_{3}+c_{0}~I=0}</math>, which confirms the above calculations.
 
This implies that theThe characteristic polynomial of matrix {{mvar|A}} is thus <math>{\displaystyle p_{A}(\lambda )=\lambda ^{3}-10\lambda ^{2}+4\lambda -40}</math>; the determinant of {{mvar|A}} is <math>{\displaystyle \det(A)=(-1)^{3} c_{0}=40}</math>; the trace is 10=−''c''<sub>2</sub>; and the inverse of {{mvar|A}} is
:<math>{\displaystyle A^{-1}=-{\frac {1}{c_{0}}}~ B_M_{3}={\frac {1}{40}} \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==