Answer
LU factorization:
$\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\3&1&0\\-\frac{1}{2}&-2&1\end{bmatrix}\begin{bmatrix}2&-4&4&-2\\0&3&-5&3\\0&0&0&5\end{bmatrix}$
Work Step by Step
To find LU factorization
$\mathbf{A}=\begin{bmatrix}2&-4&4&-2\\6&-9&7&3\\-1&-4&8&0\end{bmatrix}$
Let's Reduce the rows of the matrix;
Let $R_{2}=R_{2}-3R_{1}$
$R_{3}=R_{3}+\frac{R1}{2}$
=$\begin{bmatrix}2&-4&4&-2\\0&3&-5&3\\0&-6&10&-1\end{bmatrix}$, $K=(1,3,-\frac{1}{2})$
Let $R_{3}=R_{3}+2R_{2}$
=$\begin{bmatrix}2&-4&4&-2\\0&3&-5&3\\0&0&0&5\end{bmatrix}$,$K=(1,-2)$
Hence;
$\mathbf{U}=\begin{bmatrix}2&-4&4&-2\\0&3&-5&3\\0&0&0&5\end{bmatrix}$
To find the matrix L we replace each column by its $K$ entry;
$\mathbf{L}=\begin{bmatrix}1&0&0\\3&1&0\\-\frac{1}{2}&-2&1\end{bmatrix}$
Hence,
$\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\3&1&0\\-\frac{1}{2}&-2&1\end{bmatrix}\begin{bmatrix}2&-4&4&-2\\0&3&-5&3\\0&0&0&5\end{bmatrix}$
