Knowing inverse matrix based upon a picture?

Question

I know how to calculate an inverse matrix (by creating the augmented matrix and Gaussian elimination to get the identity matrix) and I know how to do it for a general $2 \times 2$ matrix by taking two different cases when the determinant is zero or not.

And, unless I'm mistaken, I also understand how a matrix is a linear transformation - basically a function - that can be understood as transforming the standard basis vectors. That is, I can see a simple $2 \times 2$ matrix and know what it does geometrically.

Is there a way to tell the inverse of a generic $2 \times 2$ matrix my looking at the matrix itself? I looked at Finding inverse of a matrix geometrically but I didn't get it (There were no pictures to help me gain geometric insight).

I think that I could do it for some, like the inverse of a matrix that does a stretch would be a matrix that does a compression, or a matrix that rotates counterclockwise would be a matrix that rotates clockwise -- but is there an "instant" way of knowing what the inverse matrix is based upon a picture?

Answer 1

Suppose that we are given an invertible matrix ${\bf A} \in \Bbb R^{2 \times 2}$. Suppose further that there exists a matrix ${\bf B} \in \Bbb R^{2 \times 2}$ such that $\bf B^\top$ is the left-inverse of $\bf A$, i.e.,

$$ \color{magenta}{{\bf B}}^\top \color{blue}{{\bf A}} = \begin{bmatrix} \langle \color{magenta}{{\bf b}_1} , \color{blue}{{\bf a}_1} \rangle & \langle \color{magenta}{{\bf b}_1} , \color{blue}{{\bf a}_2} \rangle \\ \langle \color{magenta}{{\bf b}_2} , \color{blue}{{\bf a}_1} \rangle & \langle \color{magenta}{{\bf b}_2} , \color{blue}{{\bf a}_2} \rangle \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

where ${\bf a}_i$ and ${\bf b}_j$ denote the $i$-th column of $\bf A$ and the $j$-th column of $\bf B$, respectively. Note that ${\bf B}^\top = {\bf A}^{-1}$. Thus,

• the 1st row of ${\bf A}^{-1}$ is orthogonal to the 2nd column of $\bf A$ and it forms an acute angle with the 1st column of $\bf A$

• the 2nd row of ${\bf A}^{-1}$ is orthogonal to the 1st column of $\bf A$ and it forms an acute angle with the 2nd column of $\bf A$

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Example

Let $ {\bf A} = \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix} $, whose inverse is $ {\bf A}^{-1} = \frac13 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} $. Pictorially,

Note that

• if the 1st column of $\bf A$ is dilated a bit, then the 1st row of $ {\bf A}^{-1} $ must contract to ensure that $\langle \color{magenta}{{\bf b}_1} , \color{blue}{{\bf a}_1} \rangle = 1$.

• if the 1st column of $\bf A$ is slightly rotated, then the 2nd row of $ {\bf A}^{-1} $ must also be rotated to ensure that $\langle \color{magenta}{{\bf b}_2} , \color{blue}{{\bf a}_1} \rangle = 0$.

Answer 2

Clearly no, because the first the question is: Is there an inverse?

While its often visible, that there is a vector maped to zero, without the determinant you are blind with respect to the question of invertibilty.

Classically, complex valued $2\times2$ matrices form a representaton of quaternions, that form not only an algebra but a field as an extension of complex 2-d numbers to 3d euclidean space.

The quaternion algebra (renamed spin by Dirac) is so complicated by its four basis matrices, that nobody wants to learn its use anymore. The rules imply all of the rot/grad/div gymnastics in electrodynamics, so that a genius like Maxwell was able to see the algebraic encoding in Hamiltons quaternions decades before the introduction of matrices and differential geometry.