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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.

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

2 Answers 2

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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.

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  • $\begingroup$ 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! $\endgroup$ Mar 11, 2019 at 17:01
  • $\begingroup$ You're very welcome. I just realized that I left out a minus sign on the second cofactor. I always make that mistake. $\endgroup$
    – saulspatz
    Mar 11, 2019 at 17:05
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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 $12*det(5) - 3*det(16) = 12*5-3*16=12$

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  • $\begingroup$ Thank you for the response. Interesting to hear that the determinant of a constant is just that constant. It certainly gives the right determinant in this case. But if this is the case, wouldn't the adjoin be that second symmetrical matrix of 60 and -48? I know this is not the correct adjugate...so what would the explanation here be? Why doesn't treating the determinant of a constant as just that constant give the correct adjugate? Note that the adjugate is the matrix of the cofactors. $\endgroup$ Mar 11, 2019 at 16:56

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