 Solution

The identity columns in the final tableau correspond to basic variables $x_2,x_3$. Their original basis matrix isalso 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 isThe last row uses the supplied convention of reduced costs $c_j-c_B^T{B}^{-1}{A}_j$, so its final right side is minus the objective. Setting the nonbasic variables to zero givesThe dual feasible vector $y=(1,1)$ has $A^{T}y=(3,4,3{)}^T\ge (1,4,3{)}^T$ and objective $4+6=10$. Part (a) independently certifies the optimum and confirms the objective-row sign.