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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?

Rodrigo de Azevedo
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Ally
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    I don't see the real interest to guess geometricaly the inverse of a matrix because there is a formula : if $$A=\pmatrix{a&b\c&d}, \ A^{-1}=\frac{1}{\det(A)}\pmatrix{d&-b\-c&a}$$. – Jean Marie Jun 06 '23 at 06:18

2 Answers2

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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 $\bf A$

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,

enter image description here

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 $\langle \color{magenta}{{\bf b}_2} , \color{blue}{{\bf a}_1} \rangle = 0$.

Rodrigo de Azevedo
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    Nice graphical explanation, in the same vein as this one that can be extended to 3D by taking $b_1$ as a certain scalar multiple of $a_2 \times a_3$ (cross product), $b_2$ as a certain scalar multiple of $a_3 \times a_1$, $b_3$ as a certain scalar multiple of $a_1 \times a_2$. – Jean Marie Jun 06 '23 at 14:37
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    ... which is illustrated in the next answer to the previous one here – Jean Marie Jun 06 '23 at 15:00
  • @JeanMarie Nice finding! – Rodrigo de Azevedo Jun 06 '23 at 15:35
  • This is great :D

    How did you think this up? (what was the inspiration)? Was it in a textbook?

     – Ally Jun 11 '23 at 21:15
  • @Ally No textbook. If one is acquainted with Gram matrices or orthogonal matrices, this is all obvious. No deep insight here. – Rodrigo de Azevedo Jun 12 '23 at 07:57
  • How do you know that acute angles are formed? I understand the orthogonal part since there is the "if and only iff" aspect of zero dot product and angle between the vectors, but how does dot product equaling $1$ indicate acute angles? (My first thought would be that the vectors would be parallel and so the angle is zero in that case?) – Ally Jun 16 '23 at 01:27
  • @Ally The angles are acute because $1 > 0$. Thus, from two choices of direction, the correct direction is fixed. All that is left to determine is the magnitude – Rodrigo de Azevedo Jun 16 '23 at 05:49
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    Okay, I suppose it's just that a positive dot product means an acute angle between the two vectors. And for your point about one's dilation and the other's contraction, that's basically, "if one is expands then the other contracts so that the dot product remains $1$". Either way, I think I've got this now and so I'll give you the bounty :D – Ally Jun 16 '23 at 22:57
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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.

Roland F
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