Is it possible to apply cofactor expansion to a $2$x$2$ matrix? Why is the determinant of a constant seemingly one?

Question

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.

Answer 1

You can do cofactor expansion of a 2x2 matrix; however, you are missing a cofactor.

The expansion would be $12*det(5) - 3*det(16)$. Determinant of a constant is just that constant so it would be $12*det(5) - 3*det(16) = 12*5-3*16=12$

Answer 2

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.