Finding eigenvector with existing eigenvalues

I have a really simple example, which I would like to use for a better understanding of Eigenvector calculation in Matlab:

A = [3 6; 4 8]

x1 = [0.75; 1]

x2 = [-2; 1]

λ1 = 11

λ2 = 0

Which code do I have to use to have the simple output of x1 & x2?

I've tried several codes I found but always got a strange output:

CODE 1:

A = [4 1; 3 2];
[V,D] = eig(A);
V1 = V(:,1)
V2 = V(:,2)

OUTPUT 1:

v1 = [0.7071; 0.7071]
v2 = [-0.3162; 0.9487]

CODE 2:

A = [4 1; 3 2];
lambda=eig(A); % you should do it by solving det(A-lambda I)=0
V = ones(2);
for k=1:2
    B = A-lambda(k)*eye(size(A));
    % select pivot column
    [~,j] = max(sum(B.^2,1));
    othercolumn = 3-j;
    V(j,k) = -B(:,j)\B(:,othercolumn);  
end
% Optional: Make eigenvectors l2 norm = 1 
V = V ./ sqrt(sum(V.^2,1))

OUTPUT 2:

V = [0.7071 -0.3162; 0.7071 0.9487]

May someone help me to get the mentioned outputs x1 = [0.75; 1] & x2 = [-2; 1]?

Thanks!!

[SALAH ALRABEEI]

you system is not linear indpendent, this means that is infinit eigenvectors. So ur x1 and x2 eig values, are just a case.

[Andreas Rüdisühli]

Oh I see, thanks!