Does rotating a matrix change its determinant?

Question

For a $2 \times 2$, it is easy to see the determinant only changes sign.

\begin{align*} \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \mapsto \left( \begin{array}{cc} c & a \\ d & b \end{array} \right) \end{align*}

We can see that $\det(A) = -\det(A')$, where $A$ is the original matrix and $A'$ is the rotated matrix. Is this always the case for any $n \times n$ matrix?

Also, this would imply that $\det(A) = \det(A'')$.

Thanks for any advice!

It's not clear what do you mean by "rotating a matrix". Can you illustrate this with a larger (such as 4x4) matrix? — user1551, Apr 26 '17 at 18:13
Rotating in the sense that if I drew the matrix on a piece of paper and turned the paper $90^\circ$ clockwise. — Enrico Borba, Apr 26 '17 at 18:14

score 5 · Accepted Answer · answered Apr 26 '17 at 17:36

5

With a $4\times 4$ matrix, rotating preserves the determinant.

In general rotating means transposing (determinant-preserving) followed by turning matrix upside down (multiplies determinant by $(-1)^{\lfloor n/2\rfloor}$).

answered Apr 26 '17 at 17:36

Angina Seng

158,341

What would multiplying by (-1)^(n/2)) mean if it was a 3x3? – Bubinga Nov 18 '21 at 06:39
There are floor brackets around the n/2 term, i.e. you round n/2 down if it’s not an integer. So for a 3x3 matrix you get a factor of -1. – John D. Cook Dec 04 '22 at 13:56

Does rotating a matrix change its determinant?

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