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Apply the LU-Algorithm to factorise the following matrix.

To factorize the matrix \( A \) using the LU algorithm, we need to decompose matrix \( A \) into the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \).

Given matrix \( A \):

\[
A = \begin{pmatrix}
-2 & 6 & 6 \\
-2 & 6 & 7 \\
-1 & 3 & 4
\end{pmatrix}
\]

Let's perform the LU factorization step-by-step:

### Step 1: Initialize \( L \) and \( U \)

We initialize \( L \) as the identity matrix and \( U \) as a copy of matrix \( A \).

\[
L = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}, \quad
U = \begin{pmatrix}
-2 & 6 & 6 \\
-2 & 6 & 7 \\
-1 & 3 & 4
\end{pmatrix}
\]

### Step 2: Perform elimination to create \( L \) and modify \( U \)

**Row 1:**

We start by using the first row of \( U \) to eliminate the below elements.

- For \( U_{21} \) (second row, first column), we need to make it zero using the first row.

\[
\text{Multiplier} = \frac{U_{21}}{U_{11}} = \frac{-2}{-2} = 1
\]

Update \( L_{21} \) and the second row of \( U \):

\[
L = \begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}, \quad
U = \begin{pmatrix}
-2 & 6 & 6 \\
0 & 0 & 1 \\
-1 & 3 & 4
\end{pmatrix}
\]

- For \( U_{31} \) (third row, first column), we make it zero using the first row.

\[
\text{Multiplier} = \frac{U_{31}}{U_{11}} = \frac{-1}{-2} = 0.5
\]

Update \( L_{31} \) and the third row of \( U \):

\[
L = \begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0.5 & 0 & 1
\end{pmatrix}, \quad
U = \begin{pmatrix}
-2 & 6 & 6 \\
0 & 0 & 1 \\
0 & 0 & 1
\end{pmatrix}
\]

**Row 2:**

- We need to eliminate \( U_{32} \) (third row, second column).

Since \( U_{22} \) is zero, and further elimination to form an upper triangular matrix is not feasible using standard LU decomposition without pivoting, we'd normally consider permutation matrices or partial pivoting in practical scenarios.

However, for the purposes of straightforward LU decomposition without pivoting:

### Step 3: Resulting \( L \) and \( U \)

The LU factorization without pivoting gives:

\[
L = \begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0.5 & 0 & 1
\end{pmatrix}, \quad
U = \begin{pmatrix}
-2 & 6 & 6 \\
0 & 0 & 1 \\
0 & 0 & 1
\end{pmatrix}
\]

### Conclusion

This is the LU decomposition of matrix \( A \) with matrices \( L \) and \( U \). Note, though, due to the presence of a zero pivot in \( U \), further row exchanges would be required for practical LU decomposition with pivoting.

Would you like to explore more about pivoting, or any specific details?

### Further Questions

1. What is the purpose of using partial pivoting in LU decomposition?
2. How can we perform LU decomposition with pivoting?
3. What are the computational complexities of LU decomposition?
4. How does LU decomposition apply to solving systems of linear equations?
5. What are some practical applications of LU decomposition in engineering and computer science?

**Tip:** When performing matrix factorizations, always consider the possibility of zero pivots and the necessity of row exchanges (pivoting) to ensure numerical stability.

Learn how to apply the LU algorithm to factorize matrix A into lower triangular matrix L and upper triangular matrix U with this detailed step-by-step guide. Includes matrix operations and LU decomposition process. Suitable for advanced mathematics enthusiasts and students.

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