I would love some direction and any information for some further reading about a cool thought I just had. If a square matrix has functions as the elements, would it not make the matrix depend on $x$ (or whatever variables) and hence have a changing determinant? What would be even cooler is for a matrix to sometimes have a determinant that is equal to $0$. Now that I think about it, this could be a rotation matrix, right? $A_{11} = \cos x$ and $A_{12} = -\sin x$ and $A_{21} = \sin x$ and $A_{22}= \cos x$ so if $x=45^\circ$, the determinant will be $0$ I think. This would mean it has no inverse for that value? I think I made a mistake because it is easy to visualize how to reverse a $45^\circ$ rotation.

In any case, please do comment anything that you know about this and where to find out more about it.

Edit: Yes the rotation matrix was a bad example because the determinant is always $1$. I could have used a better example with a determinant of $\cos x - \pi$ or $2x-6$

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Captain HD
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  • I just realized that the determinant in this case is amazingly cos^2 + sin^2 = 1 which means despite the variables, it doesn't change in this case. Ignore my 45 degree comment but I am still wondering what would have happened and why we don't let the determinant change. We could have a determinant of x, 1, 6, 2 for example which gives 2x-6 = det(A) – Captain HD Dec 26 '20 at 03:37
  • This might be related: [What's an intuitive way to think about the determinant?](https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant). – Toby Mak Dec 26 '20 at 04:01
  • The rotation matrix is actually a beautiful example of how the determinant can be constant for arbitrary values of $x$. – Toby Mak Dec 26 '20 at 04:10

2 Answers2


Yes, a matrix with functions as elements definitely has some other function as its determinant in general case. If you're interested as to where would you find such matrices without just artificially constructing one, then you could read some fluid mechanics.

For example, determinant of the strain rate tensor shows the volume rate of change of a piece of fluid at that point. You can imagine a compressible fluid flowing through a bottleneck and getting compressed along the path and that means that this tensor changes its determinant from point to point.

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The rotation matrix has a constant determinant because $\hat{i} = (\cos x, \sin x)$ and $\hat{j} = (-\sin x, \cos x)$ $ = \cos(x+\pi/2), \sin(x + \pi/2))$. The new basis vectors are just rotated by $x$ radians counterclockwise compared to the usual basis vectors $e_1 = (0, 1)$ and $e_2 = (1,0)$, hence they have the same signed area and the same determinant (for more on this, check out 3Blue1Brown's video).

The determinant can be a function, but I don't how that could be useful. Off the top of my head:

$$A = \begin{pmatrix} \cos x & \cos(-x) \\ \sin x & \sin(-x) \end{pmatrix} = \begin{pmatrix} \cos x & \cos x \\ \sin x & -\sin x \end{pmatrix}$$ has a determinant of $0$ when $x = k \pi, k \pi + \pi/2, k \in \mathbb Z$. Think about why this is in terms of the basis vectors and their span (when are they collinear?)

Toby Mak
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