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Pivot the 9 to obtain a new tableau and corresponding new solutions in bracket form

Let's tackle each part of this problem step-by-step, focusing initially on the pivot operation required for part (b). 

The tableau provided is as follows:

\[
\begin{bmatrix}
2 & 9 & 1 & 0 & 0 & 198 \\
7 & 1 & 0 & 1 & 0 & 749 \\
-10 & -6 & 0 & 0 & 1 & 0 \\
\end{bmatrix}
\]

This tableau represents a linear programming problem in the standard simplex tableau form, where the columns correspond to the variables \( x \), \( y \), \( u \), \( v \), and \( M \), and the last column represents the constants (right-hand side values).

### Step-by-Step Solution

#### Part (b) - Pivot on 9 (element in row 1, column 2)

1. **Identify the pivot element**: The pivot element is \( 9 \), located at row 1, column 2. 

2. **Row operations** to make the pivot element \( 1 \) and other elements in the pivot column \( 0 \).

   - **Row 1 (Make pivot 1)**: Divide row 1 by 9.
     \[
     R_1 = \frac{1}{9} R_1 \Rightarrow \left[\frac{2}{9}, 1, \frac{1}{9}, 0, 0, \frac{198}{9}\right] = \left[\frac{2}{9}, 1, \frac{1}{9}, 0, 0, 22\right]
     \]

   - **Row 2 (Eliminate column entry)**: Update row 2 to make the entry in the \( y \)-column (second column) zero.
     \[
     R_2 = R_2 - (1) \times R_1 = \left[7, 1, 0, 1, 0, 749\right] - \left[\frac{2}{9}, 1, \frac{1}{9}, 0, 0, 22\right]
     \]
     Simplifying,
     \[
     R_2 = \left[\frac{61}{9}, 0, -\frac{1}{9}, 1, 0, 727\right]
     \]

   - **Row 3 (Eliminate column entry)**: Update row 3 similarly to make the entry in the \( y \)-column zero.
     \[
     R_3 = R_3 + 6 \times R_1 = \left[-10, -6, 0, 0, 1, 0\right] + 6 \times \left[\frac{2}{9}, 1, \frac{1}{9}, 0, 0, 22\right]
     \]
     Simplifying,
     \[
     R_3 = \left[\frac{-10 + 12}{9}, 0, \frac{6}{9}, 0, 1, 132\right] = \left[\frac{2}{9}, 0, \frac{6}{9}, 0, 1, 132\right]
     \]

The new tableau after pivoting on the 9 will be:

\[
\begin{bmatrix}
\frac{2}{9} & 1 & \frac{1}{9} & 0 & 0 & 22 \\
\frac{61}{9} & 0 & -\frac{1}{9} & 1 & 0 & 727 \\
\frac{2}{9} & 0 & \frac{6}{9} & 0 & 1 & 132 \\
\end{bmatrix}
\]

This tableau represents the updated solution. 

Would you like me to continue with parts (a) and (c) or provide additional details on interpreting this tableau? 

---

### Follow-up Questions:

1. What are the feasible solutions of this tableau?
2. How do the group I and group II variables change after the pivot?
3. How do we determine if a solution is feasible in the simplex method?
4. What is the role of the pivot element in transforming the tableau?
5. How would another pivot on a different element affect the results?

### Tip:
When performing a pivot, always ensure the chosen element is positive and that you correctly eliminate other entries in the pivot column.

Explore the process of pivoting on element 9 in a linear programming tableau to find feasible solutions and optimize the value of M. This problem demonstrates the simplex method for transforming the tableau in linear programming, suitable for college-level learners.

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