What is the geometric intuition behind $A^{-1} = (\det A)^{-1} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$?

Question

For a 2x2 matrix $$A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$ its inverse is $$A^{-1} = (\det A)^{-1} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}.$$

This is easy to derive using Gaussian elimination and $\det A = ad - bc$. But what is the intuition behind it?

Let's write $$\tilde A :=\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$ so that $$A^{-1} = (\det A)^{-1} \tilde A.$$ Intuiting the $(\det A)^{-1}$ scalar is simple: $A^{-1}$ must undo the expansion of area of $A$, and since $\tilde A$ has the same determinant (expansion of area) as $A$, we need to divide both rows by the reciprocal of that. We use $\lambda$ such that $\det (\lambda \tilde A) = (\det A)^{-1}$, and $\det (\lambda \tilde A) = \lambda^2\det \tilde A = \lambda^2\det A$.

But what is the intuition behind $\tilde A$? Geometrically, how does $\tilde A$ relate to $A$?

Answer 1

Just expanding on my comment per request. I'll work in $V=\mathbb{R}^2$.

If $f:V\to\mathbb{R}$ is a nonzero linear functional, then $f$ can be viewed as measuring oriented lengths in $V$ (along a dual basis vector). There's a unique vector $v\in V$ such that $f$ actually measures oriented areas of parallelograms in $V$ relative to $v$: $$f(x)=\det(x,v)\tag{1}$$

Indeed, consider $D:V^2\to V$ defined by $$D(x,y)=f(x)y-f(y)x$$ Clearly $D$ is bilinear and alternating, so by the universal property of the determinant there is $v\in V$ unique with $D(x,y)=\det(x,y)v$. If $x=0$, then (1) holds trivially, and if $x\ne 0$ there is $y\in V$ with $\det(x,y)=1$, in which case $v=D(x,y)$ and hence $$\det(x,v)=\det(x,f(x)y-f(y)x)=f(x)\det(x,y)-f(y)\det(x,x)=f(x)$$ so (1) still holds.

Now given an invertible linear map (or matrix) $A:V\to V$, we have the induced nonzero linear functional $f_A=f\circ A$, and $f_A$ also measures oriented lengths. Again, there is $v'\in V$ so that $$f_A(x)=\det(x,v')$$ A very natural question is: what's the relationship between $v$ and $v'$? More specifically, when we apply $A$ before measuring lengths, how does the vector relative to which we're measuring areas change?

The answer is: $v'=\mathrm{adj}(A)v$ (or $v'=\tilde{A}v$ in your notation). In other words, $$\det(Ax,v)=\det(x,\mathrm{adj}(A)v)\tag{2}$$ and this holds for all $x,v\in V$ and fully characterizes $\mathrm{adj}(A)$.

If we take $v=Ay$ in (2), then for all $x\in V$, $$\begin{align*} \det(x,\mathrm{adj}(A)Ay)&=\det(Ax,Ay)\\ &=\det(A)\det(x,y)\\ &=\det(x,\det(A)Iy) \end{align*}$$ which implies $\mathrm{adj}(A)A=\det(A)I$ and $A^{-1}=\det(A)^{-1}\mathrm{adj}(A)$, so this is indeed the familiar adjugate.

I think this is more interesting to look at in the case of $\mathbb{R}^3$ (and higher dimension), but I leave the details of that to my video.