Vtyjkyuk

I have to find the n-th power for the following matrix $$A=beginpmatrix 1+sqrt3 & 1-sqrt3\ sqrt3 - 1 & sqrt3 +1endpmatrix
$$ My thoughts is that here could be the same situation as for $B=beginpmatrix sin x & - cos x\ cos x & sin xendpmatrix
$ which give $$B^n=beginpmatrix sin (nx) & - cos (nx) \ cos (nx) & sin (nx) endpmatrix.
$$ So I think I have to make a connection between A and B, however in A I dont have only $1$ term per place so I dont know how to proceed. $$frac12 A=beginpmatrix sin(pi /6) +cos(pi /6) &sin(pi /6) - cos(pi /6) \ cos(pi /6) - sin(pi /6) &sin(pi /6) +cos(pi /6) endpmatrix
$$ Couls you help me?

As can be easily verified
$$
left(beginarraycc
a & -b\
b & a
endarrayright)=left(beginarraycc
sintheta & -costheta\
costheta & sintheta
endarrayright)left(beginarraycc
bcostheta+asintheta & acostheta-bsintheta\
-acostheta+bsintheta & bcostheta+asintheta
endarrayright)
$$

but

$$
bcostheta+asintheta=sqrta^2+b^2left(fracbsqrta^2+b^2costheta+fracasqrta^2+b^2sinthetaright)=rhocosleft(theta_0-thetaright)
$$

and then

$$
left(beginarraycc
bcostheta+asintheta & acostheta-bsintheta\
-acostheta+bsintheta & bcostheta+asintheta
endarrayright)=rholeft(beginarraycc
cosleft(theta_0-thetaright) & sinleft(theta_0-thetaright)\
-sinleft(theta_0-thetaright) & cosleft(theta_0-thetaright)
endarrayright)
$$

so

$$
left(beginarraycc
a & -b\
b & a
endarrayright)=rholeft(beginarraycc
sintheta & -costheta\
costheta & sintheta
endarrayright)left(beginarraycc
cosleft(theta_0-thetaright) & sinleft(theta_0-thetaright)\
-sinleft(theta_0-thetaright) & cosleft(theta_0-thetaright)
endarrayright)=rholeft(beginarraycc
sintheta_0 & -costheta_0\
costheta_0 & sintheta_0
endarrayright)
$$

hence

$$
left(beginarraycc
a & -b\
b & a
endarrayright)^n=rho^nleft(beginarraycc
sinleft(ntheta_0right) & -cosleft(ntheta_0right)\
cosleft(ntheta_0right) & sinleft(ntheta_0right)
endarrayright)
$$

with

$$
theta_0 = arctanleft(frac abright)\
rho = sqrta^2+b^2
$$

Hint:
beginalign
sin(pi/6)+cos(pi/6) =& sqrt2left(fracsin(pi/6)sqrt2+fraccos(pi/6)sqrt2 right)\
=&sqrt2left(sin(pi/6)sin(pi/4)+cos(pi/6)cos(pi/4)right)=ldots
endalign

You’re on the right track by trying to pull out a common scalar factor. You just have to use a different one. Hint: A key feature of $B$ is that its rows and columns are unit vectors, so see if both rows of $A$ have the same norm. You will then have $A = kB$, from which $A^n=k^nB^n$.

$$frac12 A=beginpmatrix sin(pi /6) +cos(pi /6) &sin(pi /6) - cos(pi /6) \ cos(pi /6) - sin(pi /6) &sin(pi /6) +cos(pi /6) endpmatrix$$
$$frac12 A= beginpmatrix sin(pi /6) &cos(pi /6) \ cos(pi /6) &-sin(pi /6) endpmatrixbeginpmatrix 1 &1 \ 1 &-1 endpmatrix$$