# Multiplication

• Feb 25th 2011, 03:33 PM
lanierms
Multiplication
$a, b, c, d$ are real numbers.
When they satisfy the equations $a^2+b^2=1$, $c^2+d^2=1$, $ac+bd=0$, find the value of $ab+cd$.

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

• Feb 26th 2011, 12:03 AM
Opalg
Quote:

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

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

In terms of matrices, these equations can be written $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix }a&c\\b&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. But that implies that those two matrices are inverses of each other, so that $\begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix }a&b\\c&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. Hence $ab+cd=0.$
• Feb 26th 2011, 04:49 AM
lanierms
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.
• Feb 26th 2011, 06:03 AM
$a=-\displaytype{\frac{bd}{c}}$

$a^2+b^2=1$

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

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

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

$b^2=c^2$

I think you can use this result
• Feb 26th 2011, 06:59 PM
lanierms
Thank you so much BAdhi. I got this:

$b^2=c^2$

$b=c$ or $b=-c$

i) When $b=c$,

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

ii) When $b=-c$,

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

Is there anything wrong with it?
• Feb 26th 2011, 09:33 PM