# Let $a, b, c, d \in \mathbb R$. Given that $a^2 + c^2 = 1$, $b^2 + d^2 = 1$, and $ab + cd = 0$, show that $|ad - bc| = 1$.

As part of a proof that all 2x2 orthogonal matrices either represent rotations or reflections (Introduction to Applied Linear Algebra, exercise 10.37), I came across the below.

Let $$a, b, c, d \in \mathbb R$$. Given that $$a^2 + c^2 = 1$$, $$b^2 + d^2 = 1$$, and $$ab + cd = 0$$, I need to show that $$|ad - bc| = 1$$.

I've tried many different approaches here, but can't seem to figure out what the trick is. Any hint would be much appreciated.

Also, is there a standard approach for solving problems of this kind? If so, what is it?

• Consider $a=\cos x,c=\sin x,b=\cos y,d=\sin y,ab+cd=\cos(x-y)=0,ad-bc=-\sin(x-y)$ May 22 '20 at 21:15

Hint: Use the formula $$(a^2+c^2)(b^2+d^2)=(ad-bc)^2+(ab+cd)^2$$

$$a^2+c^2=1\implies$$ $$a=\cos(u) , c=\sin(u)$$

$$b^2+d^2=1\implies$$ $$b=\cos(v) , d=\sin(v)$$

$$ab+cd=0\implies cos(u-v)=0$$

$$\implies |\sin(u-v)|=1$$

$$\implies|\sin(u)\cos(v)-\cos(u)\sin(v)|=1$$

$$\implies |ad-bc|=1$$

• I thought of this, but it takes for granted trig identities that are not in the scope of the question, so I was looking for a starting point like the one suggested by @A. Goodier
– jeg
May 22 '20 at 21:20
• @jeg Ok.best wishes. May 22 '20 at 21:22