# Find an LU factorization of A = [(h,-4,-2,10),(h,-9,4,2),(0,0,-4,2),(0,1,4,4),(0,0,0,h/2)]. h=102

Find an LU factorization of $A=\left[\begin{array}{cccc}h& -4& -2& 10\\ h& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& \frac{h}{2}\end{array}\right]$.
h=102
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Step 1
$A=\left[\begin{array}{cccc}102& -4& -2& 10\\ 102& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& 51\end{array}\right]$
In this problem, we have to find matrices L(lower triangular) and U(upper triangular) for which A=LU
Initial Matrix: $\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}102& -4& -2& 10\\ 102& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& 51\end{array}\right]$
Using some row transformations we convert this initial matrix into LU factorization,
where L will the matrix having all diagonal entries 1, and entries above diagonal 0.
Step 2
by some row transformations, we get the matrix L and matrix U
$L=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 1& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& -\frac{1}{5}& -\frac{13}{10}& 1& 0\\ 0& 0& 0& \frac{51}{5}& 1\end{array}\right]$
$U=\left[\begin{array}{cccc}102& -4& -2& 10\\ 0& -5& 6& -8\\ 0& 0& -4& 2\\ 0& 0& 0& 5\\ 0& 0& 0& 0\end{array}\right]$
If we multiply L and U, we will get the same matrix A.