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The adjugate matrix of an $n \times n$ matrix $A$ is defined by $(\mathrm{adj}\ A)_{k\ell} = (-1)^{k+\ell}\,\det M(\ell,k)$, where $M(\ell,k)$ is the minor matrix obtained from $A$ by deleting row $\ell$ and column $k$.

The obvious application of the adjugate is the identity $A (\mathrm{adj}\ A) = (\mathrm{adj}\ A)A = (\det A)I$. Arguably this gives ample motivation, but only after a 'rabbit-from-hat' definition.

Is there a good way to motivate the adjugate matrix in a first course on matrix algebra?

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Here is one way to put the rabbit in the hat before pulling it out:

Derive the general formula for the inverse of an invertible matrix. It ends up having the form $A^{-1} = \frac{1}{\textrm{Det}(A)} \textrm{Adj}(A)$, where $\textrm{Adj}(A)$ is something we just now discovered. Hey, the formula for $\textrm{Adj}(A)$ makes sense whether $A$ is invertible or not, so it might be worth studying...

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    $\begingroup$ I feel obligated to mention here an insufficiently known fact: this formula is not how one would inverse a matrix in any case where one actually has to outside a linear algebra course, it is far too complex in term of number of operations. In fact, if one absolutely wants to compute the adjugate of a given, large invertible matrix one would be better of computing the inverse and deducing the adjugate from this formula than computing $n^2$ determinants. $\endgroup$
    – Benoît Kloeckner
    Mar 19, 2016 at 14:31
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    $\begingroup$ @BenoîtKloeckner Of course I agree. Is there any place where we use determinants in an undergraduate course where it would not be computationally more efficient to use some other tool? I suppose the change of variables formula is one, but Cramer's rule, finding eigenvalues, etc are all better done without. $\endgroup$
    – Steven Gubkin
    Mar 19, 2016 at 22:21
  • $\begingroup$ @BenoîtKloeckner Not sure if this is really "using" the determinant, but how about finding the area of a region after a transformation? Unless you already know the formula for the area of an ellipse, it is nice to derive it by knowing that the area of the final region is just the original area multiplied by the determinant of the matrix of the transformation. $\endgroup$
    – Nick C
    Nov 14, 2017 at 20:43
  • $\begingroup$ @NickC I agree with Steven and you, the determinant really is useful as a volume (or volume distortion, be it for affine transformations or differentiable maps). It is its raison d'être, I would say. $\endgroup$
    – Benoît Kloeckner
    Nov 17, 2017 at 18:41
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Slight variant on Steven's answer: You don't actually have to derive the general formula. Make them work out the inverse of $\left( \begin{smallmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{smallmatrix} \right)$. (if they don't have it memorized already). Point out that they could clearly do $3 \times 3$ and $4 \times 4$ if they worked harder. Pass out paper with the $3 \times 3$ and $4 \times 4$ results printed on it ("I'll save you the time"). Ask them what patterns they see.

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    $\begingroup$ I like this variation since it mitigates the problem that the $2 \times 2$ case is too small to really see what's going on, whereas the $3 \times 3$ case is already big enough to require significant calculation. $\endgroup$
    – Mark Wildon
    Mar 24, 2016 at 18:21
  • $\begingroup$ This is a great suggestion! $\endgroup$
    – Steven Gubkin
    Mar 24, 2016 at 18:48

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