To 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!

Re: 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).

Re: 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 D² = 1.

thus D = det(A) = ±1.

now for:

$A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$

we have:

$A^{-1} = \frac{1}{\det(A)}\begin{bmatrix}d&-b\\-c&a \end{bmatrix}}$

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:

$A = \begin{bmatrix}a&b\\b&-a \end{bmatrix}$ .

since det(A) = -1 = -a² - b², it follows that:

a² + b² = 1, or equivalently, b = ±√(1 - a²). thus A is of the form:

$\begin{bmatrix}a&\pm \sqrt{1-a^2}\\ \pm\sqrt{1-a^2}&-a \end{bmatrix}$ for $a \in [-1,1]$ (the signs on the square root terms must match).

if we so desired, we could write this as:

$A = \begin{bmatrix}\cos(\theta)&\sin(\theta)\\ \sin(\theta)&-\cos(\theta)\end{bmatrix}$ for $0 \leq \theta < 2\pi$

which is a REFLECTION about the line with slope: $\tan\left(\frac{\theta}{2}\right)$ .

ignore what chiro said. there's only two rotations that work, but infinitely many reflections.

Re: 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].