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!

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- Feb 10th 2013, 09:46 PMJordi1986To find all invertible matrices of A
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!

- Feb 11th 2013, 06:18 PMchiroRe: To find all invertible matrices of A
Hey Jordi1986.

Hint: Consider the orthogonal matrices which have a determinant of 1 (look up rotation matrices if you are still confused). - Feb 12th 2013, 12:42 AMDevenoRe: To find all invertible matrices of A
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. - Feb 12th 2013, 02:00 AMchiroRe: To find all invertible matrices of A
You're right I didn't see the A = A^(-1) [which does restrict the solution space a lot more].