= Solution
{c}
The identity columns in the final tableau correspond to basic variables $x_2,x_3$. Their original basis matrix is
$$
B=\begin{pmatrix}2&1\\2&2\end{pmatrix},\qquad
B^{-1}=\begin{pmatrix}1&-1/2\\-1&1\end{pmatrix},
$$
also visible directly in the two slack columns. Multiply the original constraint tableau and right side by $B^{-1}$ to obtain the missing entries. Subtract four times the first row and three times the second from the initial objective row, using the basic costs. The completed <simplex tableau> is
$$
\boxed{\begin{array}{rrrrr|r}
3/2&1&0&1&-1/2&1\\
-1&0&1&-1&1&2\\ \hline
-2&0&0&-1&-1&-10
\end{array}.}
$$
The last row uses the supplied convention of reduced costs $c_j-c_B^TB^{-1}A_j$, so its final right side is minus the objective. Setting the nonbasic variables to zero gives
$$
\boxed{(x_1,x_2,x_3)=(0,1,2),\qquad f_{\max}=10.}
$$
The dual feasible vector $y=(1,1)$ has $A^Ty=(3,4,3)^T\geq(1,4,3)^T$ and objective $4+6=10$. Part (a) independently certifies the optimum and confirms the objective-row sign.
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