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I know, that the variance of OLS $\beta = \sigma^2 (X^TX)^{-1}$. Then I did calculations of $ (X^TX)^{-1}$ and I need the inverse in order to complete my proof. But the inverse of 3*3 matrix is a pain(especially in general case).

So I need a hint.Is there another way to prove the statement above? Or the only option is taking painful inverse of 3*3 matrix?

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One hint is that $N^{-1} \sum_i (X_i^TX_i)$ converges to $\mathbb{E}(X^TX)$, which is almost the variance. Recall that in the scalar case, $V(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$. And since your variables are demeaned, you know something about their expectation.

Next, $X$ has three components: the first is a constant so it has zero variance and zero covariance with the other two. So there are only four relevant entries in the $3\times 3$ matrix. And when you know the correlation, $r$, you can get a little further.

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