You can derive the formula for the determinant

Wednesday September 15, 2021

“Determinants are difficult, non-intuitive, and often defined without motivation.” (Sheldon Axler, 1995)

It's common to talk about properties of the determinant, but then treat its formula as almost coming from nowhere, focusing more on mnemonics than meaning. I think it could be worth seeing that the determinant's formula pops out easily in a common setting.

A matrix is a transformation that maps origin to origin. If a matrix happens to map anything else to the origin, then that matrix isn't invertible, because no inverse can map the origin back to two different points. So we seek a solution to a system of equations.

\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Eliminating \( x \) and \( y \) (example solution below) yields the familiar \( ad - bc = 0 \). That is, if \( ad - bc = 0 \), there's some \( x \) and \( y \) (not both zero) that the matrix maps to the origin, and so the matrix is not invertible. The determinant is \( ad - bc \).

I worked through this also for the three-by-three case and was satisfied to recover the usual expression. There's probably a deeper proof or other connection to more linear algebra, but I don't recall ever being taught any rationale at all for the determinant taking the form it does, so I was just pleased that it's as straightforward as this to show at this level. It's slower, but if you happen to forget the formula for the determinant, you can always work it out again this way!

Example solution

The matrix equation above is equivalent to Equations 1 and 2.

\[ ax + by = 0 \tag{1} \]

\[ cx + dy = 0 \tag{2} \]

Solving Equation 1 for \( x \) produces Equation 3.

\[ x = - \frac{b}{a} y \tag{3} \]

Substituting Equation 3 into Equation 2 yields Equation 4.

\[ - \frac{cb}{a}y + dy = 0 \tag{4} \]

Multiplying by \( a \), dividing by \( y \), and reordering then recovers the familiar form of the determinant in Equation 5.

\[ ad - bc = 0 \tag{5} \]

If I was just sleeping through linear algebra, please let me know what I missed! Also let me know if there are any problems with or alternatives to this method! Thanks!

The quote at top from Axler is from his 1995 paper, Down with Determinants!.

Thanks to Erica for helpful feedback!