Prove $cof(A^t) = cof(A)^t$ I'm trying to use $A^{-1} = cof(A)^tD$, where $D = det(A)^{-1}$ to prove $cof(A^t) = cof(A)^t$. I end up with statements these two $A^{-1} = Dcof(A)^t$. And $(A^{t})^{-1} = D cof(A^t)$. But I don't know how to use this to show: $cof(A^t) = cof(A)^t$.

Let $d=\det A = \det(A^t)$. Then

$$cof(A)^t = d \cdot A^{-1} =d \cdot \big( (A^{-1})^t\big)^t = \big( d \cdot( A^{-1})^t \big)^t = \big(d \cdot( A^t)^{-1}\big)^t = cof(A^t)$$

In that fourth equality, we used the fact that $(A^{-1})^t=(A^t)^{-1}$.