Looking for confirmation: Let A be 3x3 matrix and det(A)=2 then
$$det(-2A) = ?$$ $$det(\frac 1 2 * A^{-1}) = ?$$ $$det((2A)^3) = ?$$ $$det(3A^T)^{-1} = ?$$
So
$$det(-2A) = -2^3*det(A) = -8*2 = -16$$ => Using property $$det(kA)=k^{n}*det(A)$$ $$det(\frac 1 2A^{-1}) = \frac 1 2^{3}*\frac 1 {det(A)} = \frac 1 2^{3}*\frac 1 2 = \frac 1 8*\frac 1 2 = \frac 1 {16} $$ => Using property $$|A^{-1}| = \frac 1{|A|} $$ $$det(2A^{3}) = 2^3*det(A^3) = 2^3*det(A)^3=8*2^3= 8*8 = 64$$ Using property $$detA^{k+1}=det(A^k)*det(A)=(detA)^k*det(A)=det(A)^k+1$$