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Let's solve part (c), which asks for the LU factorization of matrix \( A \) using **full pivoting** based on the matrix provided:

\[
A = \begin{bmatrix}
1 & -3 & 1 \\
2 & 3 & 6 \\
-1 & 2 & 2
\end{bmatrix}
\]

Using full pivoting, which means that we can interchange both rows and columns, the LU factorization process will be carried out with the following steps:

### Step 1: Initial Setup and Permutation

In full pivoting, we look for the largest absolute element in the matrix and pivot both the row and column. The largest element in this case is \( 6 \), located at position \( (2,3) \). So we swap rows and columns to bring it to the top left:

\[
A = \begin{bmatrix}
6 & 3 & 2 \\
0 & -7/2 & 2/3 \\
0 & 0 & -31/21
\end{bmatrix}
\]

With the permutation matrices applied as well, we have the following LU decomposition:

\[
P = \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]

and

\[
Q = \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]

### Step 2: Gaussian Elimination and LU Decomposition

Now, perform Gaussian elimination to transform \( A \) into upper triangular form:

1. **Step 1**: Eliminate below the pivot \( 6 \):

   Divide row 1 by 6 and subtract from row 2 to make the element below 6 zero:

   \[
   L = \begin{bmatrix}
   1 & 0 & 0 \\
   1/6 & 1 & 0 \\
   1/3 & -2/7 & 1
   \end{bmatrix}
   \]
   
   After this step, \( U \) becomes:

   \[
   U = \begin{bmatrix}
   6 & 3 & 2 \\
   0 & -7/2 & 2/3 \\
   0 & 0 & -31/21
   \end{bmatrix}
   \]

### Conclusion:

The LU factorization of \( A \) with full pivoting is:

\[
A = P L U Q^T
\]

Where:

- \( P \) is the row permutation matrix,
- \( L \) is the lower triangular matrix,
- \( U \) is the upper triangular matrix,
- \( Q^T \) is the transpose of the column permutation matrix.

### Summary of Matrices:
- **Permutation matrices**:
  \[
  P = Q = \begin{bmatrix}
  0 & 1 & 0 \\
  1 & 0 & 0 \\
  0 & 0 & 1
  \end{bmatrix}
  \]
  
- **Lower triangular matrix**:
  \[
  L = \begin{bmatrix}
  1 & 0 & 0 \\
  1/6 & 1 & 0 \\
  1/3 & -2/7 & 1
  \end{bmatrix}
  \]
  
- **Upper triangular matrix**:
  \[
  U = \begin{bmatrix}
  6 & 3 & 2 \\
  0 & -7/2 & 2/3 \\
  0 & 0 & -31/21
  \end{bmatrix}
  \]

Would you like more details on any part of the calculation?

### Related Questions:

1. How does full pivoting affect numerical stability in LU factorization?
2. What is the purpose of permutation matrices in LU decomposition?
3. Can you explain the difference between partial and full pivoting?
4. What are the key benefits of rook pivoting compared to partial pivoting?
5. How can LU factorization be used to solve systems of linear equations?

### Tip:
When performing LU factorization with full pivoting, always ensure you carefully track both row and column swaps to keep the permutation matrices consistent.

Now solve part (c) using full pivoting based on the provided matrix.

Learn how to compute the LU factorization of a matrix using full pivoting. This example walks through the complete Gaussian elimination process and includes the usage of permutation matrices for accurate factorization.

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