计算行列式
解:
|a`````1`````0`````0|
|-1````b`````1`````0|
|0````-1`````c`````1|
|0`````0````-1`````d|
上式=(r1+ar2)=
|0````1+ab````a`````0|
|-1`````b`````1`````0|
|0`````-1`````c`````1|
|0``````0````-1`````d|
=[r1+(ab+1)r3]=
|0```````0```c(ab+1)+a``ab+1|
|-1``````b```````1`````````0|
|0``````-1```````c`````````1|
|0``
`````0``````-1`````````d|
=[r1+(abc+a+c)r4]=
|0```````0```````0```ab+1+d(abc+a+c)|
|-1``````b```````1``````````0```````|
|0``````-1```````c``````````1```````|
|0```````0``````-1``````````d```````|
=按(1,4)元展开, 得
abcd+ad+cd+ab+1。
。
[展开]
方法一:由行列式展开定理,按2*2分块,前两列只有选1,2行并对应后两列选3,4行,以及前两列选1,3行并对应后两列选2,4行,其余余子式积都是零。结果得 a*d +(ab+1)*(cd+1) = abcd+ab+cd+ad+1,注意代数符号。
方法二:代入Mathematica验证,Det[{{a,1,0,0},{-1,b,1,0},{0,-1,c,1},{0,0,-1,d}}] = abcd + ab + cd +1,结果一致。
abcd+cd+ab+ad+1,