# find [MATH]\alpha , \beta[/MATH]

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• May 3rd 2009, 10:01 PM
zorro
find [MATH]\alpha , \beta[/MATH]
If A=$\displaystyle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, find $\displaystyle \alpha$,$\displaystyle \beta$ so that $\displaystyle (\alpha l + \beta A)^2 = A$
• May 4th 2009, 01:23 AM
Gamma
I think that is supposed to be the Identity matrix?

So you need
$\displaystyle \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2$

Well multiply it out and see what you get. What I got was:
$\displaystyle \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta$
and
$\displaystyle 2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2$

But then use 1 and we see we need $\displaystyle \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$
• Dec 23rd 2009, 12:14 AM
zorro
I got this ?
Quote:

Originally Posted by Gamma
I think that is supposed to be the Identity matrix?

So you need
$\displaystyle \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2$

Well multiply it out and see what you get. What I got was:
$\displaystyle \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta$
and
$\displaystyle 2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2$

But then use 1 and we see we need $\displaystyle \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$

$\displaystyle ( \alpha I + \beta A)^2$ = $\displaystyle \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}$

Is this right and what should i do next ????
• Dec 23rd 2009, 03:43 AM
HallsofIvy
Quote:

Originally Posted by zorro
$\displaystyle ( \alpha I + \beta A)^2$ = $\displaystyle \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}$

Is this right and what should i do next ????

According to the time stamp, gamma gave you the answer 8 months ago. Why are you asking now?