<s>[INST] Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where 
$ A =
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}
$, then $A =\begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
    a & c \\
    b & d
  \end{bmatrix}
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}$ = $\begin{bmatrix}
    a^2 + c^2 & ab + cd \\
    ab + cd & b^2 + d^2
  \end{bmatrix}$ = $\begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle . 
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
 [/INST] from $A^TA = I$, we konw $(a, c)$ and $(b, d)$ is orthogonal, for some $\theta$, we know $(a, c) = \sqrt{(a^2 + c^2)}(\cos\theta, \sin\theta) = (\cos\theta, \sin\theta)$, the unit vector orthogonal to $(a, c)$ is $(-\sin\theta, \cos\theta)$, so we know, for some scalar $k$, $(b, d) = k(-\sin\theta, \cos\theta)$. $k$ is either $1$ or $-1$. since $\det(A) = 1$, 
here we used $\sqrt{a^2 + c^2} = 1$, and $\sqrt{b^2 + d^2} = 1$
Let us calculate
$$\det(A) = \det \begin{bmatrix} \cos\theta & \sin\theta \\  -k\sin\theta & k\cos\theta \end{bmatrix} = k(\cos^2\theta + \sin^2\theta) = k = 1$$
so 
$$ A = \begin{bmatrix} \cos\theta & \sin\theta \\  -\sin\theta & \cos\theta \end{bmatrix}$$
</s>