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:<math>M_3= A^2-A\mathrm{tr} A -\frac{1}{2}\Bigl(\mathrm{tr} A^2 -(\mathrm{tr} A)^2\Bigr) I, \qquad c_{n-3}=- \tfrac{1}{6}\Bigl( (\operatorname{tr}A)^3-3\operatorname{tr}(A^2)(\operatorname{tr}A)+2\operatorname{tr}(A^3)\Bigr)=-\frac{1}{3}(c_n \mathrm{tr} A^3+c_{n-1} \mathrm{tr} A^2 +c_{n-2}\mathrm{tr} A); </math>
etc.,<ref>Zadeh, Lotfi A. and Desoer, Charles A. (1963, 2008). ''Linear System Theory: The State Space Approach'' (Mc Graw-Hill; Dover Civil and Mechanical Engineering) ISBN 9780486466637</ref>
Observe {{math|''A<sup>−1</sup> {{=}} − M<sub>n</sub> /c<sub>0</sub>'' {{=}} (−)<sup>''n''−1</sup>''M<sub>n</sub>''/det''A''}} terminates the recursion at {{mvar| λ}}. This could be used to obtain the inverse or the determinant of {{mvar|A}}.
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