两个公式

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第一个公式：

这个公式的证明需要用到

第二个公式：

证明如下：

[egin{array}{l}
left| {{I_{p imes p}} + {C_{p imes n}}A_{n imes n}^{ - 1}{B_{n imes p}}} ight| = left| {egin{array}{*{20}{c}}
{{I_{p imes p}} + {C_{p imes n}}A_{n imes n}^{ - 1}{B_{n imes p}}}&0\
0&{{I_n}}
end{array}} ight|\
mathop {{ m{ = = = = = = = = = = }}}limits^{youchengleft| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&0\
{{B_{n imes p}}}&{{I_n}}
end{array}} ight|} left| {egin{array}{*{20}{c}}
{{I_{p imes p}} + {C_{p imes n}}A_{n imes n}^{ - 1}{B_{n imes p}}}&0\
{{B_{n imes p}}}&{{I_n}}
end{array}} ight|\
mathop {{ m{ = = = = = = = = = = = }}}limits^{zuochengleft| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&{ - {C_{p imes n}}A_{n imes n}^{ - 1}}\
0&{{I_n}}
end{array}} ight|} left| {egin{array}{*{20}{c}}
{{I_{p imes p}} + {C_{p imes n}}A_{n imes n}^{ - 1}{B_{n imes p}} - {C_{p imes n}}A_{n imes n}^{ - 1}{B_{n imes p}}}&{ - {C_{p imes n}}A_{n imes n}^{ - 1}}\
{{B_{n imes p}}}&{{I_n}}
end{array}} ight|{ m{ = }}left| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&{ - {C_{p imes n}}A_{n imes n}^{ - 1}}\
{{B_{n imes p}}}&{{I_n}}
end{array}} ight|\
mathop {{ m{ = = = = = = = = = = = }}}limits^{youchengleft| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&{{C_{p imes n}}A_{n imes n}^{ - 1}}\
0&{{I_n}}
end{array}} ight|} left| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&{{C_{p imes n}}A_{n imes n}^{ - 1} - {C_{p imes n}}A_{n imes n}^{ - 1}}\
{{B_{n imes p}}}&{{I_n}{ m{ + }}{B_{n imes p}}{C_{p imes n}}A_{n imes n}^{ - 1}}
end{array}} ight|{ m{ = }}left| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&0\
{{B_{n imes p}}}&{{I_n}{ m{ + }}{B_{n imes p}}{C_{p imes n}}A_{n imes n}^{ - 1}}
end{array}} ight|\
{ m{ = }}left| {egin{array}{*{20}{c}}
{{I_{p imes p}}}&0\
{{B_{n imes p}}}&{{I_n}{ m{ + }}{B_{n imes p}}{C_{p imes n}}}
end{array}} ight| cdot left| {A_{n imes n}^{ - 1}} ight|{ m{ = }}left| {A_{n imes n}^{ - 1}} ight| cdot left| {{A_{n imes n}} + {B_{n imes p}}{C_{p imes n}}} ight|
end{array}]