If det(A)=k\det(A) = kdet(A)=k, then det(kA)\det(kA)det(kA) for an n×nn \times nn×n matrix is:
kn+1k^{n+1}kn+1
kndet(A)k^n \det(A)kndet(A)
k⋅det(A)k \cdot \det(A)k⋅det(A)
knk^nkn