Why does $A^{-1} = \frac{1}{|A|} \cdot \operatorname{adj}(A)$ work, intuitively?

Question

Why does:

$A^{-1} = \frac{1}{|A|} \cdot \operatorname{adj}(A)$

work?

I've looked but I can't find anything which is annoying because there definitely is something.

Anyway, thanks in advance! Cheers!

EDIT: https://www.youtube.com/watch?v=_mvsgdfz-Ik This video explains it perfectly!

$2^{nd}$ EDIT: I realize that I don't understand why the adjugate works so perfectly there

Answer 1

Yay! I figured it out! I was just trying to simply write it out, and I got it.

It can all be seen in this screen shot:

Answer 2

You can easily derive this formula for 2x2 and 3x3 matrices if you find $A^{-1}$ by row reducing $A$ in an augmented matrix with the identity matrix on the right hand block of the augmented matrix.

Answer 3

If you're willing to generalize the cross-product a little bit, it's not so tough. I'm going to say that the cross product for vectors in $\Bbb R^k$ takes $k-1$ vectors and produces a $k$th one. For $k = 3$ (the familiar case), the cross product consumes two vectors and produces a third one. For $k = 2$, it consumes just $1$ vector, and the "formula" is $$ \times \pmatrix{a\\b} = \pmatrix{b\\a} $$

The thing to recognize is that for any vector $v \in \Bbb R^2$, $v \cdot (\times v)$ is zero: the cross product is perpendicular to the "input" vector.

This generalizes to $\Bbb R^3$: the cross product $ u \times v$ is perpendicular to both $u$ and $v$. And this works in general: in $\Bbb R^k$, if $$ u_k = v_1 \times \ldots \times v_{k-1}, $$ then $u_k$ is perpendicular to each $v_i$. (The reason I've chosen to call this $u_k$ will be apparent soon).

Of course, I have to give you a formula for this generalized cross-product before any of this makes sense, but let me proceed for a moment. Suppose we transpose $u_k$ to get a row vector, and matrix-multiply it by each of the $v_i$. We find that $$ [u_k^t] [v_i] = 0, $$ right? Because this matrix product, of a $k \times 1$ vector by a $1 \times k$ vector, amounts to just a dot-product. Now suppose that we have one last vector, $v_k$, that we didn't include when we computed $u_k$. What do we get when we make a $k \times k$ matrix whose columns are the $v_i$, and multiply it by $u_k^t$? \begin{align} u_k^t [v_1, v_2, \ldots, v_k] &= [u_k\cdot v_1, \ldots, u_k \cdot v_k]\\ &= [0, \ldots,0, u_k \cdot v_k], \end{align} which is almost the last row of the identity matrix.

Well, what if we define $u_1$ by taking the cross-product of all the $v$-vectors except $v_1$, and $u_2$ by taking the cross-products of all the $v$-vectors except $v_2$, and so on? Then we end up with \begin{align} \pmatrix{ u_1^t\\ u_2^t\\ \cdots\\ u_k^t} \pmatrix{v_1, v_2, \ldots, v_k} = \pmatrix{ u_1 \cdot v_1 & 0 & 0 & \ldots & 0 \\ 0 & u_2 \cdot v_2 & 0 & \ldots & 0 \\ \ldots\\ 0 & 0 & 0 & \ldots & u_k \cdot v_k } \end{align} and we've almost found the inverse of the matrix $\pmatrix{v_1, v_2, \ldots, v_k}$!

You might try using the formula for the cross product of one vector in $\Bbb R^2$ to see what "almost inverse" you get for the $2 \times 2$ matrix $\pmatrix{a & b \\ c & d}$.

But back to that big formula. If we could arrange for each of those diagonal entries $u_i \cdot v_i$ to be the same number, we could just divide through by that number and we'd have found the inverse. And Cramer's rule shows that the number we should be aiming for is the determinant of the matrix whose columns are all the $v_i$s.

This leads me to my "definition" of the cross product: the cross product of $v_1, \ldots v_{k-1}$ (if they're linearly independent) is the unique vector $u$ such that

1. $u$ is orthogonal to each of the $v_i$, and

2. $\det \pmatrix{v_1, v_2, \ldots, v_{k-1}, u } = \det \pmatrix{v_1, v_2, \ldots, v_{k-1}, v_k}$.

(When the $v$'s are linearly dependendent, their cross product is defined to be zero).

Now you still have to figure out why the sums-of-products formula for the determinants used in the cofactors is what you need to give an explicit formula for the cross-product ... but that's not really the tough part: all you have to do is show that that sum has the two properties described in items 1 and 2 above, which follow from elementary properties of the determinant.