Hey I have covariance matrix:
$$C=\begin{pmatrix} 0,01 & 0.01 & 0\\ \\ 0.01 & 0,02 & -0.01 \\ \\ 0 & -0.01 & 0,03 \end{pmatrix}$$
So the variance of porfolio is:
$$\sigma_w^2=\begin{pmatrix}w_1 & w_2 & w_3 \end{pmatrix} \begin{pmatrix}0.01 & 0.01 & 0\\0.01 & 0.02 & -0.01\\0 & -0.01 & 0.03 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$$
And i want to find portoflios on MVL wchich standard deviation is $$\sigma_w=\frac{1}{10}$$. IS it possible to find this porfolios having only this information? 
!
EDIT:
I tried to do it using the Lagrange multiplier method, which gave the same result, I don't know why it works :( Here's what I do: I use lagrange multipliers to minimalize function $f(w_1,w_2,w_3)=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$ under condition $g_1(w_1,w_2,w_3)=w_1+w_2+w_3-1=0$ I get equation system $\left \{\begin{array}{lr}0.02w_1+0.02w_2-\lambda=0\\0.04w_2+0.02w_1-0.02w_3-\lambda=0 \\0.06w_3-0.02w_2-\lambda=0 \\w_1+w_2+w_3-1=0 \\\end{array} \right.$ I know that if I solved it to the end I would get a wallet with minimal variance, but I don't want to do it, so not using the equation of 3 gets: $-0.02w_2+0.02w_3=0 \Rightarrow w_2=w_3$ (from 1 and 2) and $w_1=1-2w_2$ from 4 now I put these for equation $\sigma_{w}^{2}=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$ where $ \sigma_{w}= \frac{1}{10}$ and solve it $0.01=0.01(1-2w_2)^2+0.02w_2^2+0.03w_2^2+0.02w_2(1-2w_2)-0.02w_2^2$ Finally I get 2 solutions $w_1=(-\frac{1}{3},\frac{2}{3},\frac{2}{3})$ and $w_2=(1,0,0)$ which are correct.My question is why it works? Can anyone explain me?
I didn't solve the equations to the end, I just determined relationships between the weights and just put in a specific variance. Why can I do it, and if I can't, why did I get a good result?
