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 $|ad - bc| = 1$.

Question

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?

Answer 1

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

Answer 2

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

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

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

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

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

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