设Aij为aij的代数余子式.
把行列式按第一行展开,有det(A)=a11*A11+a12*A12+a13*A13
因为aij=Aij,故det(A)=(a11)^2+(a12)^2+(a13)^2
又因为aij=Aij,所以有：
a11=a22a33-a32a23.(1)
a22=a11a33-a31a13.(2)
a33=a11a22-a21a12.(3)
a32=a21a13-a11a23.(4)
a23=a31a12-a11a32.(5)
将(2)至(5)式带入(1)式,有
a11=(a11a33-a31a13)(a11a22-a21a12)-(a21a13-a11a23)(a31a12-a11a32)
将右侧展开,整理后有：
a11=(a11)^2(a22a33-a23a32)+a11a13(a21a32-a22a31)+a11a12(a23a32-a21a33)
因为
a11=a22a33-a23a32,a13=a21a32-a22a31,a12=a23a32-a21a33
所以
a11=(a11)^3+a11(a13)^2+a11(a12)^2
有因为a11≠0,两边同除以a11有：
1=(a11)^2+(a12)^2+(a13)^2
即：det(A)=(a11)^2+(a12)^2+(a13)^2=1
所以行列式为1