# Find the LU-Factorization of the matrix A below A=[(2,1,-1),(-2,0,3),(2,1,-4),(4,1,-4),(6,5,-2)]

Find the LU-Factorization of the matrix A below
$A=\left[\begin{array}{ccc}2& 1& -1\\ -2& 0& 3\\ 2& 1& -4\\ 4& 1& -4\\ 6& 5& -2\end{array}\right]$
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Step 1
Given
$\left[\begin{array}{ccc}2& 1& -1\\ -2& 0& 3\\ 2& 1& -4\\ 4& 1& -4\\ 6& 5& -2\end{array}\right]$
Step 2
Solution
L=Lower triangular matrix
U=Upper triangular matrix
$A=\left[\begin{array}{ccc}2& 1& -1\\ -2& 0& 3\\ 2& 1& -4\\ 4& 1& -4\\ 6& 5& -2\end{array}\right]\sim \left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 2\\ 0& 0& -3\\ 0& -1& -2\\ 0& 2& 1\end{array}\right]\left(\begin{array}{c}{R}_{2}\to {R}_{2}+{R}_{1}\\ {R}_{3}\to {R}_{3}-{R}_{1}\\ {R}_{4}\to {R}_{4}-2{R}_{1}\\ {R}_{5}\to {R}_{5}-3{R}_{1}\end{array}\right)$
$\sim \left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 2\\ 0& 0& -3\\ 0& 0& 0\\ 0& 0& -3\end{array}\right]{R}_{4}\to {R}_{4}+{R}_{2},{R}_{5}\to {R}_{5}-2{R}_{2}$
$\sim \left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 2\\ 0& 0& -3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]{R}_{5}\to {R}_{5}-{R}_{3}$
$U=\left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 2\\ 0& 0& -3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$
Step 3
And lower triangular matrix that is L is calculated by the entries are the number in the 0 respectively.
i.e., $L=\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 1& 0& 1\\ 2& -1& 0\\ 3& 2& 1\end{array}\right]$