There exist two involutions on ring $B$ as follows:

$\begin{pmatrix}
a&b\\
0&c
\end{pmatrix}^{\ast_1}= \begin{pmatrix}
c^{\ast}&b^{\ast}\\
0&a^{\ast}
\end{pmatrix},$
and
$\begin{pmatrix}
a&b\\
0&c
\end{pmatrix}^{\ast_2}= \begin{pmatrix}
c^{\ast}&-b^{\ast}\\
0&a^{\ast}
\end{pmatrix}$.

An involution is proper if $aa^{\ast}=0$, then $a=0$. If the $\ast$-ring $R$ is Baer $\ast$, then $\ast$ is a involution proper.

The involutions $\ast_1$ and $\ast_2$ are not proper, because
$\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}^{\ast_1}=\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}^{\ast_2}=\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}\begin{pmatrix}
0&0\\
0&1
\end{pmatrix}=0$ and $\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}\not=0$.
Hence $B=T_2(A)$ is not a Baer $\ast$-ring.