一般地,对于行列式
Dn=|abcabc⋅⋅⋅⋅bca|
存在
Dn=aDn−1−bcDn−2
其中:
D1=aD2=|abca|D3=|abcabca|以此类推
Related Issues not found
Please contact @kb1000fx to initialize the comment
Related Issues not found
Please contact @kb1000fx to initialize the comment
v1.5.2