Non-real numbers in system of equations Given $a^2+b^2=1$,
$c^2+d^2=1$,
$ac+bd=0$
To prove
$a^2+c^2=1$,$b^2+d^2=1$,$ab+cd=0$
Now this can be easily done by trigonometric substitution if it was given that $a,b,c,d$ are real numbers.
I have a solution using matrix which I think is valid even if the given numbers are non real.
https://artofproblemsolving.com/community/c6h475289p2661966
I am curious to know if there other ways to it Algebraical or with any other method.
This says that
$$\pmatrix{a&amp;b\\c&amp;d}\pmatrix{a&amp;c\\b&amp;d}=\pmatrix{1&amp;0\\0&amp;1}$$
implies
$$\pmatrix{a&amp;c\\b&amp;d}\pmatrix{a&amp;b\\c&amp;d}=\pmatrix{1&amp;0\\0&amp;1}.$$
This follows from the fact that (over any commutative ring) any square matrix
with a left inverse has a right inverse (necessarily the same as the left inverse).
But if you want an &quot;equational&quot; proof avoiding concepts from linear algebra
observe that
$$a^2+c^2=1+a^2c^2-(1-a^2)(1-c^2)=a+a^2c^2-b^2d^2=1+(ac+bd)(ac-bd)=1$$
and similarly for $b^2+d^2=1$. Also
$$ab+cd=ab(c^2+d^2)+(a^2+b^2)cd=(ac+bd)(ad+bc)=0.$$
The vectors $(a,b)$ and $(c,d)$ are unit and orthogonal.
A reflection around the bissector of the first quadrant swaps the abscissas and the ordinates, and creates the unit orthogonal vectors $(a,c)$ and $(b,d)$.


By trigonometry, the general solution of the first system is
$$a=\cos\theta,b=\sin\theta,c=\pm\sin\theta,d=\mp\cos\theta.$$