1. ## Multiplication

$\displaystyle a, b, c, d$ are real numbers.
When they satisfy the equations $\displaystyle a^2+b^2=1$, $\displaystyle c^2+d^2=1$, $\displaystyle ac+bd=0$, find the value of $\displaystyle ab+cd$.

I got as far as getting $\displaystyle a^2+c^2=1$, but I can't seem to go any further.

2. Originally Posted by lanierms
$\displaystyle a, b, c, d$ are real numbers.
When they satisfy the equations $\displaystyle a^2+b^2=1$, $\displaystyle c^2+d^2=1$, $\displaystyle ac+bd=0$, find the value of $\displaystyle ab+cd$.

I got as far as getting $\displaystyle a^2+c^2=1$, but I can't seem to go any further.

In terms of matrices, these equations can be written $\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix }a&c\\b&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. But that implies that those two matrices are inverses of each other, so that $\displaystyle \begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix }a&b\\c&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. Hence $\displaystyle ab+cd=0.$

3. Aren't there any other ways than using matrix? Because I didn't learn matrix yet and I have no idea what you are saying.

4. $\displaystyle a=-\displaytype{\frac{bd}{c}}$

$\displaystyle a^2+b^2=1$

$\displaystyle \displaytype{\frac{b^2d^2}{c^2}}+b^2=1$

$\displaystyle b^2(d^2+c^2)=c^2$

since $\displaystyle d^2+c^2=1$

$\displaystyle b^2=c^2$

I think you can use this result

5. Thank you so much BAdhi. I got this:

$\displaystyle b^2=c^2$

$\displaystyle b=c$ or $\displaystyle b=-c$

i) When $\displaystyle b=c$,

$\displaystyle ac+bd=ab+bd=ab+cd=0$

ii) When $\displaystyle b=-c$,

$\displaystyle ac+bd=-ab+bd=-ab-cd=-(ab+cd)=0$

Is there anything wrong with it?

6. just the way i thought