Find an \(LU\)-factorization of \(A\).
\[A=\begin{bmatrix}5 & -5 & 10 & 0 & 5\\-3 & 3 & 2 & 2 & 1\\-2 & 2 & 0 & -1 & 0\\1 & -1 & 10 & 2 & 5\end{bmatrix}\]
\[A=LU=\begin{bmatrix}\answer {5} & 0 & 0 & 0\\\answer {-3} & \answer {8} & 0 & 0\\\answer {-2} & \answer {4} & \answer {-2} & 0\\\answer {1} & \answer {8} & \answer {0} & \answer {1}\end{bmatrix}\begin{bmatrix}\answer {1} & \answer {-1} & \answer {2} & \answer {0} & \answer {1}\\0 & 0 & \answer {1} & \answer {1/4} & \answer {1/2}\\0 & 0 & 0 & \answer {1} & \answer {0}\\0 & 0 & 0 &0 & 0\end{bmatrix}\]

Source

[Nicholson] W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2021, Open Edition, Example 2.7.3.