Hint: Consider the orthogonal matrices which have a determinant of 1 (look up rotation matrices if you are still confused).
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.
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 AT = A-1 = A, which means A must be symmetric, so we have b = c.
thus our matrix is of the form:
since det(A) = -1 = -a2 - b2, it follows that:
a2 + b2 = 1, or equivalently, b = ±√(1 - a2). thus A is of the form:
for (the signs on the square root terms must match).
if we so desired, we could write this as:
which is a REFLECTION about the line with slope: .
ignore what chiro said. there's only two rotations that work, but infinitely many reflections.