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

Question
Polynomial factorization
asked 2021-02-13
Find an LU factorization of \(\displaystyle{A}={\left[\begin{array}{cccc} {h}&-{4}&-{2}&{10}\\{h}&-{9}&{4}&{2}\\{0}&{0}&-{4}&{2}\\{0}&{1}&{4}&{4}\\{0}&{0}&{0}&\frac{{h}}{{2}}\end{array}\right]}\).
h=102

Answers (1)

2021-02-14
Step 1
\(\displaystyle{A}={\left[\begin{array}{cccc} {102}&-{4}&-{2}&{10}\\{102}&-{9}&{4}&{2}\\{0}&{0}&-{4}&{2}\\{0}&{1}&{4}&{4}\\{0}&{0}&{0}&{51}\end{array}\right]}\)
In this problem, we have to find matrices L(lower triangular) and U(upper triangular) for which A=LU
Initial Matrix: \(\displaystyle{\left[\begin{array}{ccccc} {1}&{0}&{0}&{0}&{0}\\{0}&{1}&{0}&{0}&{0}\\{0}&{0}&{1}&{0}&{0}\\{0}&{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{0}&{1}\end{array}\right]}{\left[\begin{array}{cccc} {102}&-{4}&-{2}&{10}\\{102}&-{9}&{4}&{2}\\{0}&{0}&-{4}&{2}\\{0}&{1}&{4}&{4}\\{0}&{0}&{0}&{51}\end{array}\right]}\)
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
\(\displaystyle{L}={\left[\begin{array}{ccccc} {1}&{0}&{0}&{0}&{0}\\{1}&{1}&{0}&{0}&{0}\\{0}&{0}&{1}&{0}&{0}\\{0}&-\frac{{1}}{{5}}&-\frac{{13}}{{10}}&{1}&{0}\\{0}&{0}&{0}&\frac{{51}}{{5}}&{1}\end{array}\right]}\)
\(\displaystyle{U}={\left[\begin{array}{cccc} {102}&-{4}&-{2}&{10}\\{0}&-{5}&{6}&-{8}\\{0}&{0}&-{4}&{2}\\{0}&{0}&{0}&{5}\\{0}&{0}&{0}&{0}\end{array}\right]}\)
If we multiply L and U, we will get the same matrix A.
0

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