Let's stick with the nomenclature in the literature and let $\gamma$ denote the decision maker's risk aversion coefficient. The optimization problem is
$$
\max_{\mathrm{w}} \mathrm{w}^T\mathrm{\mu}-\frac{1}{2}\gamma \mathrm{w}^T\mathrm{\Sigma}\mathrm{w} \quad s.t. \mathrm{w}^T\mathrm{e}=1
$$
where $e$ denotes a vector of ones. The corresponding Lagrangean reads:
$$
L(\mathrm{w},\lambda)= \mathrm{w}^T\mathrm{\mu}-\frac{1}{2}\gamma \mathrm{w}^T\mathrm{\Sigma}\mathrm{w} -\lambda\left( \mathrm{w}^T\mathrm{e}-1\right)
$$
The first order conditions are linear:
\begin{align}
L_w&=\mathrm{\mu}-\gamma\mathrm{\Sigma}\mathrm{w}-\lambda\mathrm{e}=!=0\\
L_{\lambda}&=\mathrm{e}^T\mathrm{w}=!=1
\end{align}
We can formulate this as a linear system as:
\begin{equation}
\begin{pmatrix}
\gamma\mathrm{\Sigma} & \mathrm{e}\\
\mathrm{e}^T & 0\end{pmatrix}
\begin{pmatrix}
\mathrm{w}\\
\lambda\end{pmatrix}=\begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix}
\end{equation}
and hence
\begin{equation}
\begin{pmatrix}
\mathrm{w^*}\\
\lambda^*\end{pmatrix}=
\begin{pmatrix}
\gamma\mathrm{\Sigma} & \mathrm{e}\\
\mathrm{e}^T & 0\end{pmatrix}^{-1}
\begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix}=\begin{pmatrix}c_{11} & c_{12} \\ c_{21} & c_{22}\end{pmatrix}\begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix}
\end{equation}
This is where we use the block matrix inversion theorem . We know that $w^*$ is given by the first row of the inverted matrix times the constraint vector,
$$
\mathrm{w}^*=c_{11}\mathrm{\mu}+c_{12}
$$
Looking up the two inversions from wiki, we find
$$
c_{11}=\frac{1}{\gamma}\mathrm{\Sigma}^{-1}-\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\left(\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\right)^{-1}\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1}
$$
and
$$
c_{12}=\frac{1}{\gamma}\mathrm{\Sigma^{-1}}\mathrm{e}\left(\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\right)^{-1}=\frac{\mathrm{\Sigma^{-1} e}}{\mathrm{e}^T\mathrm{\Sigma^{-1} e}}
$$
Let us introduce the following
\begin{align}
a&=\mathrm{e}^T\mathrm{\Sigma^{-1}}\mathrm{e}\\
b&=\mathrm{e}^T\mathrm{\Sigma^{-1}}\mathrm{\mu}\\
w_{MVP}&=\frac{\mathrm{\Sigma^{-1} e}}{\mathrm{e}^T\mathrm{\Sigma^{-1} e}}\\
w_{Tangency}&=\frac{\mathrm{\Sigma^{-1} \mu}}{\mathrm{e}^T\mathrm{\Sigma^{-1} \mu}}
\end{align}
Note that, conveniently, the two portfolios are the minimum-variance portfolio and the tangency portfolio. Then, $c_{12}\mu$ simplifies to
$$
c_{12}\mathrm{\mu}=\frac{1}{\gamma}bw_{Tangency}-\frac{1}{\gamma}w_{MVP}b
$$
and thus
$$
w^* = w_{MVP} + \frac{1}{\gamma}b\left(w_{Tangency}-w_{MVP}\right)
$$
Finally, we note that $E(MVP)=\frac{b}{a}$, and $V(MVP)=\frac{1}{a}$. Thus, we may ultimately replace $b$ in equation above and arrive at
$$
w^* = w_{MVP} + \frac{1}{\gamma}\frac{\mu_{MVP}}{\sigma_{MVP}^2}\left(w_{Tangency}-w_{MVP}\right)
$$
