计算行列式:a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/21 20:44:58
计算行列式:a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d
a 1 0 0 ;
-1 b 1 0;
0 -1 c 1;
0 0 -1 d;
a 1 0 0 ;
-1 b 1 0;
0 -1 c 1;
0 0 -1 d;
a,1,0,0
-1,b,1,0
0,-1,c,1
0,0,-1,d
r2+(1/a)r1:
a,1,0,0
0,(ba+1)/a,1,0
0,-1,c,1
0,0,-1,d
r3+[a/(ab+1)]r2:
a,1,0,0
0,(ba+1)/a,1,0
0,0,(cba+c+a)/(ba+1),1
0,0,-1,d
r4+[(ba+1)/(cba+c+a)]r3:
a,1,0,0
0,(ba+1)/a,1,0
0,0,(cba+c+a)/(ba+1),1
0,0,0,(dcba+dc+da+ba+1)/(cba+c+a)
=a*[(ba+1)/a]*[(cba+c+a)/(ba+1)]*[(dcba+dc+da+ba+1)/(cba+c+a)]
=ab+ad+cd+abcd+1
-1,b,1,0
0,-1,c,1
0,0,-1,d
r2+(1/a)r1:
a,1,0,0
0,(ba+1)/a,1,0
0,-1,c,1
0,0,-1,d
r3+[a/(ab+1)]r2:
a,1,0,0
0,(ba+1)/a,1,0
0,0,(cba+c+a)/(ba+1),1
0,0,-1,d
r4+[(ba+1)/(cba+c+a)]r3:
a,1,0,0
0,(ba+1)/a,1,0
0,0,(cba+c+a)/(ba+1),1
0,0,0,(dcba+dc+da+ba+1)/(cba+c+a)
=a*[(ba+1)/a]*[(cba+c+a)/(ba+1)]*[(dcba+dc+da+ba+1)/(cba+c+a)]
=ab+ad+cd+abcd+1
一个四阶行列式谁来教教我吧...计算下列行列式:a 1 0 0-1 b 1 00 -1 c 10 0 -1 d答案是 a
求四阶行列式 a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d
行列式 a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d
四阶行列式求解!a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d
这种行列式的值怎么求a 1 0 0-1 b 1 00 -1 c 10 0 -1 d
计算行列式:|2 1 0 0 0|
行列式的计算 1 1 1 1 a b c d a^2 b^2 c^2 d^2 a^4 b^4 c^4 d^4
计算行列式D={1 a bc}{1 b ca}{1 c ab}的值 D=(b-a)(b-c)(a-c)怎么算
求4阶行列式 a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d 硬算怎么算?
a 1 0 0 -1 b 1 0 0 -1 c 1 0 0 -1 d 帮我解一下这个行列式
行列式的计算(过程)(1) (2)a 1 0 0 1+x 1 1 1-1 b 1 0 1 1-x 1 10 -1 c 1
设a,b,c是互异的实数,则行列式 1 1 1 a b c a^3 b^3 c^3 =0 的充要条件?