Proof: Adding additional regressor and the influence on the adjusted R^2 I'm looking at the influence of an additional regressor in an OLS-model and on the adjusted $\bar{R}^2$. What I have to proove is that $\bar{R}^2$ rises if and only if the square of the respective t-statistic is bigger than 1. I found a solution to the proof in "Greene - Econometric Analysis, (Chapter 3, exercise 9)" and managed to replicate the steps more or less to get to the final result, which is as follows:
$\frac{b_k^2(x_k'M_1x_k)}{(s^2)}>1$,
where $b_k$ is the coefficient of the additional regressor in the long model and $s^2$ is its estimated variance. In my understanding the squared t-statistic of this regressor should just be 
$\frac{b_k^2}{(s^2)}$.
How do I interpret the rest of the nominator $(x_k'M_1x_k)$? Is this even the right proof I'm looking for? 
The complete solution I'm looking at is given in this PDF (p5, ex. 9):
pages.stern.nyu.edu/~wgreene/Text/Greene_6e_Solutions_Manual.pdf
 A: Recall that the t-statistic on the $k$-th coefficient (for testing $\beta_k=0$ assuming homoskedasticity) in a linear regression is given by
$$
t=\frac{b_k}{\sqrt{s^2(X'X)^{-1}_{kk}}},
$$
where $(X'X)^{-1}_{kk}$ denotes the $k$-th diagonal element of $(X'X)^{-1}$, so that the squared t-ratio is
$$
t=\frac{b_k^2}{s^2(X'X)^{-1}_{kk}}.
$$
This is the same as your
$$
\frac{b_k^2(x_k'M_1x_k)}{s^2}=\frac{b_k^2}{s^2/(x_k'M_1x_k)}
$$
due to the Frisch-Waugh-Lovell theorem discussed in for example Chapter 3 of Greene's textbook, or more specifically/directly, as an application of the partitioned inverse lemma:
For $A\;(m \times m)$, $B\;(m \times n)$, $C\;(n \times m)$ and $D\;(n \times n)$ we have that, provided the respective inverses exist,
$$
\left(%
\begin{array}{cc}
A & B \\
C & D \\
\end{array}%
\right)^{-1}=
\left(%
\begin{array}{cc}
A^{-1} + A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}
& -A^{-1}B(D-CA^{-1}B)^{-1} \\
-(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\
\end{array}%
\right).
$$
For $A=X_1'X_1$ (with $X_1$ the other regressors except $x_k$), $B=X_1'x_k$, $C=x_k'X_1$ and $D=x_k'x_k$, we can plug in for the lower-right element of the inverse to directly obtain that 
$$
(D-CA^{-1}B)^{-1}=(X'X)^{-1}_{kk}=1/(x_k'M_1x_k).
$$ 
