Solved: the determinant of a constant has the same value as that constant. It is possible to apply cofactor expansion to a two-by-two matrix.
For example, this matrix:
$$ \left[
\begin{array}{cc}
12&3\\
16&5\\
\end{array}
\right] $$
The cofactor expansion would be $12*det(5)$, seeing as taking out the first row and column leaves just $[5]$. Likewise, the other cofactors would be: $-3det(16), -16det(3), $ and $5det(12)$.
It would seem that the determinant of any constant is $1$. I say this because the adjugate of the above matrix is not
$$ \left[
\begin{array}{cc}
60&-48\\
-48&60\\
\end{array}
\right] $$
which is what it would be if the determinant of a constant had the same value as the constant. Instead the adjugate/ajoint is:
$$ \left[
\begin{array}{cc}
12&-3\\
-16&5\\
\end{array}
\right] $$
which only makes sense if the determinant of a constant is $1$. (I know this is the ajoint because it gives the correct inverse when multiplied by $1/12$)
So is the answer to my first question yes? Can you apply cofactor expansion to a $2$x$2$ matrix? If so, is there an explanation for why the determinant of a constant is equal to $1$?
Edit: Apologies, I realized my adjugate is wrong. Since this is the case, even my assertion that $\operatorname{det} k=1$ of a constant $k$ does not have any ground.
Assuming the determinant of a constant k is: $\operatorname{det} k = k$, the first adjoint matrix I wrote (the one consisting of two $60$s and two $-48$s) would be the correct adjoint. But this is clearly not the adjoint. I remain confused.
Best Answer
You've got the wrong idea about what the cofactor is. In the cofactor expansion $$\det\begin{bmatrix}12&3\\16&5\end{bmatrix}=12\det(5)-3\det(16),$$ the cofactors are $\det(5)$ and $-\det(16)$ The coefficients $12$ and $3$ are not part of the cofactors. There's a detailed discussion here.