Answer
The LU factorization:
$\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\\frac{1}{2}&1&0\\-3&-2&0\end{bmatrix}\begin{bmatrix}2&-4&2\\0&7&-5\\0&0&0\end{bmatrix}$
Work Step by Step
To find LU factorization of the matrix
$\mathbf{A}=\begin{bmatrix}2&-4&2\\1&5&-4\\-6&-2&4\end{bmatrix}$
Let's Reduce the rows of the matrix;
Let $R_{2}=R_{2}-\frac{R_{1}}{2}$
=$\begin{bmatrix}2&-4&2\\0&7&-5\\-6&-2&4\end{bmatrix}$
Let $R_{3}=R_{3}+3R_{1}$
=$\begin{bmatrix}2&-4&2\\0&7&-5\\0&-14&10\end{bmatrix}$
Let $R_{2}=R_{3}-2R_{2}$
=$\begin{bmatrix}2&-4&2\\0&7&-5\\0&0&0\end{bmatrix}$
Hence;
$\mathbf{U}=\begin{bmatrix}2&-4&2\\0&7&-5\\0&0&0\end{bmatrix}$
To find the matrix L we divide each column by its pivot entry;
$\mathbf{L}=\begin{bmatrix}\frac{2}{2}&0&0\\\frac{1}{2}&\frac{7}{7}&0\\-\frac{6}{2}&-\frac{14}{7}&0\end{bmatrix}=\begin{bmatrix}1&0&0\\\frac{1}{2}&1&0\\-3&-2&0\end{bmatrix}$
Hence,
$\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\\frac{1}{2}&1&0\\-3&-2&0\end{bmatrix}\begin{bmatrix}2&-4&2\\0&7&-5\\0&0&0\end{bmatrix}$
