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?

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

score 10 · Accepted Answer · answered May 22 '20 at 21:14

10

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

answered May 22 '20 at 21:14

A. Goodier

10,964

1

This is exactly the type of hint I needed. I'll accept when it lets me :) – jeg May 22 '20 at 21:19
Glad I could help :) – A. Goodier May 22 '20 at 21:21
1

https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity – qwr May 23 '20 at 05:25

score 1 · Answer 2 · answered May 22 '20 at 21:17

1

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

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

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

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

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

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

answered May 22 '20 at 21:17

hamam_Abdallah

62,951

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. – hamam_Abdallah May 22 '20 at 21:22

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 $|ad - bc| = 1$.

2 Answers2