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Matrix Inversion question?
Ok, I need some help with a question:
Find all invertible matrices A of the form
A= a b
c d (square matrix)
and satisfying A = A^-1 and A^T = A^-1.
They also provide a hint: the identity cos^2(t) + sin^2(t) = 1may be useful.
I had a go at it but didn't get a particularly useful answer.
1 Answer
- kbLv 78 years agoFavorite Answer
Since we want A = A^(-1) = A^t:
Taking determinants yields |A| = 1/|A| = |A| ==> |A| = ±1.
So, if A =
[a b]
[c d], then
A^t =
[a c]
[b d], and
A^(-1) = ±
[d -b]
[-c a].
---------------
(i) If |A| = 1:
Then A = A^(-1) ==> d = a, b = -b and c = -c
and A^t = A^(-1) ==> a = d, and c = -b
Thus, we need a = d and b = c = 0.
Since |A| = ad - bc = 1, we have that a = d = ±1.
Hence, A = ±I.
----------------
(ii) If |A| = -1:
Then A = A^(-1) ==> a = -d, b = b, c = c,
and A^t = A^(-1) ==> a = -d, b = c.
So, A is of the form
[a b]
[b -a], with -a^2 - b^2 = -1 <==> a^2 + b^2 = 1.
We can parameterize this via a = cos t and b = sin t.
So, we have A =
[cos t sin t]
[sin t -cos t].
I hope this helps!
