= Solution
With nonnegative production amounts $x_1,x_2,x_3$, the primal linear programme is
$$
\boxed{\max 5x_1+6x_2+3x_3,\qquad4x_1+2x_2\leq6,\quad2x_1+3x_2+2x_3\leq11.}
$$
Introduce slacks $s_R,s_S\geq0$. The initial basic variables are the slacks, and the reduced profits of the goods are $5,6,3$. The greatest is six, so $x_2$ enters. The <ratio test> compares $6/2=3$ and $11/3$, so $s_R$ leaves. The resulting dictionary is
$$
x_2+2x_1+\tfrac12s_R=3,\qquad
s_S-4x_1+2x_3-\tfrac32s_R=2,
$$
with profit $18-7x_1+3x_3-3s_R$. The greatest positive reduced profit is now three, so $x_3$ enters; its only positive pivot coefficient is two in the second row, and $s_S$ leaves. The final tableau, expressed as equations, is
$$
\begin{aligned}
x_2+2x_1+\tfrac12s_R&=3,\\
x_3-2x_1-\tfrac34s_R+\tfrac12s_S&=1,\\
P&=21-x_1-\tfrac34s_R-\tfrac32s_S.
\end{aligned}
$$
All nonbasic reduced profits are nonpositive, proving optimality at
$$
\boxed{(x_1,x_2,x_3)=(0,3,1),\qquad P=21.}
$$
The ordered optimal <basis> is $(x_2,x_3)$, hence
$$
\boxed{B=\begin{pmatrix}2&0\\3&2\end{pmatrix},\qquad
B^{-1}=\begin{pmatrix}1/2&0\\-3/4&1/2\end{pmatrix}.}
$$
The inverse is read from the two slack columns of the final basic equations.
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