= Solution
Subtracting the second row from the third gives a third row $(0,2,0)$, so expansion along it shows $\det B=-2\ne0$. Thus the <matrix inverse> exists and every right-hand side has a unique solution.
To calculate it, temporarily solve $B\mathbf x=\mathbf b$ with general $\mathbf b=(b_1,b_2,b_3)^T$. Subtracting the second equation from the third and from the first gives
$$
2x_2=b_3-b_2,\qquad x_1-x_2=b_1-b_2.
$$
Back-substitution into the second equation gives
$$
x_1=b_1-\frac32b_2+\frac12b_3,\quad x_2=-\frac12b_2+\frac12b_3,\quad x_3=-b_1+\frac72b_2-\frac32b_3.
$$
Therefore
$$
\boxed{B^{-1}=\begin{pmatrix}1&-3/2&1/2\\0&-1/2&1/2\\-1&7/2&-3/2\end{pmatrix}.}
$$
For the given right-hand side, <matrix multiplication> yields the unique answer
$$
\boxed{\mathbf x=B^{-1}(1,2,3)^T=(-1/2,1/2,3/2)^T.}
$$
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