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