计算行列式计算一个四阶行列式：第一行|a,1,0,0|，第二行|-1,b,1,0|，第三行，|0,-1,c,1|第四行|0,0,-1,d|。请教详细解题过程个借题思路。谢谢！
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````````...全部
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。
收起
