Question

求行列式的值 D=1 3 0 1 3 0 1 4 1 1 2 1 0 1 1 1 求过程解答谢谢

Answer 1

A31+3A32-2A33+2A34=|(3,1,-1,2)(-5,1,3,-4)(1,3,-2,2)(1,-5,3,-3)|【将r3的各元素替换成(1,3,-2,2)】=|(3,1,-1,2)(-5,1,3,-4)(1,3,-2,2)(5/2,-1/2,0,0)|【r4+r3*(3/2)】=|(8,1,-1,2)(0,1,3,-4)(16,3-2,2)(0,-1/2,0,0)|【c1+c2*5】=(-1/2)*|(8,-1,2)(0,3,-4)(16,-2,2)|【按r4】=(-8)*|(1,-1,1)(0,3,-2)(2,-2,1)|【提出c1、c3公因子】=(-8)*|(1,-1,1)(0,3,-2)(0,0,-1)|【r3+r1*(-2)】=(-8)*|(3,-2)(0,-1)|【按c1】=(-8)*[3*(-1)-(-2)*0]=24

Answer 2