This will now be my 22nd post on linear algebra, and I hope it’ll be noticeable, looking at all of them together so far, that we haven’t talked about systems of equations. And there’s a good reason for that: because they suck you down an ugly hole of mindless calculation, meaning-challenged tedium, and fruitless, pointless backward thinking. Systems are awesome and important, and I’m sure someone can make a strong argument for introducing them early, but, for me, they can wait.

The inverses of matrices are pretty interesting. The inverse of a 2 × 2 matrix is the matrix that, when multiplied to the original matrix, gives the product that is the identity matrix. The inverse is given by the middle matrix below: \[\begin{bmatrix}\mathtt{a} & \mathtt{b}\\\mathtt{c} & \mathtt{d}\end{bmatrix}\begin{bmatrix}\mathtt{\,\,\,\,\frac{d}{ad-bc}} & \mathtt{-\frac{b}{ad-bc}}\\\mathtt{-\frac{c}{ad-bc}} & \mathtt{\,\,\,\,\frac{a}{ad-bc}}\end{bmatrix}\mathtt{=}\begin{bmatrix}\mathtt{1} & \mathtt{0}\\\mathtt{0} & \mathtt{1}\end{bmatrix}\]

There is an easy-to-follow derivation of this formula here if you’re interested—one that requires only some high school algebra.

You can notice, given this setup, that the identity matrix is its own inverse. Are there any others that fit this description? One thought: we need \(\mathtt{ad-bc}\) to be equal to 1, and we also want \(\mathtt{a=d}\). The values on the other diagonal, \(\mathtt{b}\) and \(\mathtt{c}\), should both be 0, since they have to be equal to their opposites.

In that case, \(\begin{bmatrix}\mathtt{-1} & \mathtt{\,\,\,\,0}\\\mathtt{\,\,\,\,0} & \mathtt{-1}\end{bmatrix}\) would be its own inverse too. What does that mean?

For the identity matrix, it means that if we apply the do-nothing transformation to the home-base matrix, we stay on home base, with the “x” vector pointed to (1, 0) and the “y” vector pointed to (0, 1). The matrix above represents a reflection across the origin. So, applying a reflection across the origin to it should return us to home base.

Wouldn’t a reflection across the x-axis be an inverse of itself, then? Yes! The criteria above for an inverse need to be amended a little to allow for negatives effectively canceling each other out to make positives. \[\begin{bmatrix}\mathtt{1} & \mathtt{\,\,\,\,0}\\\mathtt{0} & \mathtt{-1}\end{bmatrix}\begin{bmatrix}\mathtt{1} & \mathtt{\,\,\,\,0}\\\mathtt{0} & \mathtt{-1}\end{bmatrix}\mathtt{=}\begin{bmatrix}\mathtt{1} & \mathtt{0}\\\mathtt{0} & \mathtt{1}\end{bmatrix}\]

The inverse of a matrix is represented with a superscript \(\mathtt{-1}\). So, the inverse of the matrix \(\mathtt{A}\) is written as \(\mathtt{A^{-1}}\).

The Transpose

The transpose of a matrix is the matrix you get when you take the rows of the matrix and turn them into columns instead. The transpose of a matrix is represented with a superscript \(\mathtt{T}\). So: \[\begin{bmatrix}\mathtt{\,\,\,\,1} & \mathtt{3}\\\mathtt{-2} & \mathtt{4}\end{bmatrix}^{\mathtt{T}}\mathtt{=}\begin{bmatrix}\mathtt{1} & \mathtt{-2}\\\mathtt{3} & \mathtt{\,\,\,\,4}\end{bmatrix}\]

If a 2 × 2 matrix, \(\mathtt{V}\) is orthogonal—meaning that its column vectors are perpendicular and each column vector has a length of 1—then its transpose is the same as its inverse, or \(\mathtt{V^{-1}=V^T}\).