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

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.

• The cofactor matrix is the matrix obtained by replacing each element by its cofactor, NOT by the product of the element and its cofactor. Mar 11, 2019 at 17:03

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.
• OH MY GOODNESS. That makes complete sense now. Yes the element $a_{ij}$ is not part of the cofactor $C_{ij}$. Everything has fallen into place, thank you so much! Mar 11, 2019 at 17:01
The expansion would be $$12*det(5) - 3*det(16)$$. Determinant of a constant is just that constant so it would be $$12*det(5) - 3*det(16) = 12*5-3*16=12$$