1

I've seen Why is the determinant of a rotation matrix equal to 1?

I do get that no matter what value you put for theta you will end up getting $1$, but I would like to explain it using the cos and sin graphs (do some reasoning based on graphs).

I think there might be a way to proof this using the graphs.

Thanks.

TQFT
  • 323
zаѓатhᵾѕтѓа
  • 37
  • 1
    Do you want to use a graph to show that the determinant of $\left( \begin{matrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{matrix} \right)$ is $1$? I suppose you could plot $y=\cos^2(\theta)+\sin^2(\theta)$ – Henry Dec 13 '21 at 01:50
  • @Henry How would that show that the determinant is 1? – zаѓатhᵾѕтѓа Dec 13 '21 at 01:54
  • 4
    Have you taken trigonometry? By the way, @Henry you have a typo in the second column of your matrix. The minus sign should go with the sine, not the cosine. – John Douma Dec 13 '21 at 02:08
  • 4
    When you rotate an object, its image is congruent to its original. Think about that for a moment – imranfat Dec 13 '21 at 02:32
  • 1
    Well! I believe you know $\theta$ matrix but that is not what you want but the geometrical/intuitional thought for the same, is that so? – Darshan P. Dec 13 '21 at 06:43

1 Answers1

1

The general form of a $2 \times 2$ rotation matrix is:

$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}$

By definition, the determinant of this matrix is $\cos^2{\theta} + \sin^2{\theta}$. This is equal to $1$ by the Pythagorean trigonometric identity.

Allure
  • 616