Well, we should be pretty comfortable moving things around with vectors and matrices. We’re good on some of the forward thinking. We can think of a matrix \(\mathtt{A}\) as a mapping of one vector (or an entire set of vectors) to another vector (or to another set of vectors). Then we can think of \(\mathtt{B}\) as the matrix which undoes the mapping of \(\mathtt{A}\). So, \(\mathtt{B}\) is the inverse of \(\mathtt{A}\).

How do we figure out what \(\mathtt{A}\) and \(\mathtt{B}\) are?

\[\mathtt{B}\color{green}{\begin{bmatrix}\mathtt{-4} \\\mathtt{\,\,\,\,\,1} \end{bmatrix}} = \color{green}{\begin{bmatrix}\mathtt{\,\,3\,\,} \\\mathtt{\,\,3\,\,} \end{bmatrix}}\]

Eyeballing Is a Lost Art in Mathematics Education

It is! We can figure out the matrix \(\mathtt{A}\) without doing any calculations. Break down the movement of the green point into horizontal and vertical components. Horizontally, the green point is reflected across the “y-axis” and then stretched another third of its distance from the y-axis. This corresponds to multiplying the horizontal component of the green point by –1.333…. For the vertical component, the green point starts at 3 and ends at 1, so the vertical component is dilated by a factor of 0.333…. We can see both of these transformations shown in the change in the sizes and directions of the blue and orange basis vectors. So, our transformation matrix \(\mathtt{A}\) is shown below. When we multiply the vector (3, 3) by this transformation matrix, we get the point, or position vector, (–4, 1). \[\begin{bmatrix}\mathtt{-\frac{4}{3}} & \mathtt{0}\\\mathtt{\,\,\,\,0} & \mathtt{\frac{1}{3}}\end{bmatrix}\begin{bmatrix}\mathtt{3}\\\mathtt{3}\end{bmatrix} = \begin{bmatrix}\mathtt{-4}\\\mathtt{\,\,\,\,1}\end{bmatrix}\]

You can see that \(\mathtt{A}\) is a scaling matrix, which is why it can be eyeballed, more or less. And what is the inverse matrix? We can use similar reasoning and work backward from (–4, 1) to (3, 3). For the horizontal component, reflect across the y-axis and scale down by three fourths. For the vertical component, multiply by 3. So, the inverse matrix, \(\mathtt{B}\), when multiplied to the vector, produces the correct starting vector: \[\begin{bmatrix}\mathtt{-\frac{3}{4}} & \mathtt{0}\\\mathtt{\,\,\,\,0} & \mathtt{3}\end{bmatrix}\begin{bmatrix}\mathtt{-4}\\\mathtt{\,\,\,\,1}\end{bmatrix} = \begin{bmatrix}\mathtt{3}\\\mathtt{3}\end{bmatrix}\]

You’ll notice that we use the reciprocals of the non-zero scaling numbers in the original matrix to produce the inverse matrix. You can do the calculations with the other points on the animation above to test it out.

Incidentally, we can also eyeball the eigenvectors—those vectors which don’t change direction but are merely scaled as a result of the transformations—and even the eigenvalues (the scale factor of each transformed eigenvector). The vector (1, 0) is an eigenvector, with an eigenvalue of \(\mathtt{-\frac{4}{3}}\) for the original transformation and an eigenvalue of –0.75 for the inverse, and the vector (0, 1) is an eigenvector, with an eigenvalue of \(\mathtt{\frac{1}{3}}\) for the original transformation and an eigenvalue of 3 for the inverse.