I am asked to find all the invertible matrices of the form A=[(a, b) (c, d)] that satisfy A=A^-1 and A^t=a^-1 any cluues? I just know that the identity cos^2t+sin^2t=1 may be useful, cheeers!
since A = A^{-1}, and:
det(A^{-1}) = 1/det(A), we must have:
det(A) = D = 1/det(A) = 1/D, so D^{2} = 1.
thus D = det(A) = ±1.
now for:
$\displaystyle A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$
we have:
$\displaystyle A^{-1} = \frac{1}{\det(A)}\begin{bmatrix}d&-b\\-c&a \end{bmatrix}}$
if det(A) = 1, this leads to b = c = 0, a = d = ±1, that is: A = ±I (both of these matrices fit the bill).
the case det(A) = -1 is more interesting:
here, we can only conclude d = -a.
however, we know by the orthogonality of A, that A^{T} = A^{-1} = A, which means A must be symmetric, so we have b = c.
thus our matrix is of the form:
$\displaystyle A = \begin{bmatrix}a&b\\b&-a \end{bmatrix}$.
since det(A) = -1 = -a^{2} - b^{2}, it follows that:
a^{2} + b^{2} = 1, or equivalently, b = ±√(1 - a^{2}). thus A is of the form:
$\displaystyle \begin{bmatrix}a&\pm \sqrt{1-a^2}\\ \pm\sqrt{1-a^2}&-a \end{bmatrix}$ for $\displaystyle a \in [-1,1]$ (the signs on the square root terms must match).
if we so desired, we could write this as:
$\displaystyle A = \begin{bmatrix}\cos(\theta)&\sin(\theta)\\ \sin(\theta)&-\cos(\theta)\end{bmatrix}$ for $\displaystyle 0 \leq \theta < 2\pi$
which is a REFLECTION about the line with slope: $\displaystyle \tan\left(\frac{\theta}{2}\right)$.
ignore what chiro said. there's only two rotations that work, but infinitely many reflections.