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)]

Question
Polynomial factorization
asked 2021-02-24
Find the LU-Factorization of the matrix A below
\(\displaystyle{A}={\left[\begin{array}{ccc} {2}&{1}&-{1}\\-{2}&{0}&{3}\\{2}&{1}&-{4}\\{4}&{1}&-{4}\\{6}&{5}&-{2}\end{array}\right]}\)

Answers (1)

2021-02-25
Step 1
Given
\(\displaystyle{\left[\begin{array}{ccc} {2}&{1}&-{1}\\-{2}&{0}&{3}\\{2}&{1}&-{4}\\{4}&{1}&-{4}\\{6}&{5}&-{2}\end{array}\right]}\)
Step 2
Solution
L=Lower triangular matrix
U=Upper triangular matrix
\(\displaystyle{A}={\left[\begin{array}{ccc} {2}&{1}&-{1}\\-{2}&{0}&{3}\\{2}&{1}&-{4}\\{4}&{1}&-{4}\\{6}&{5}&-{2}\end{array}\right]}\sim{\left[\begin{array}{ccc} {2}&{1}&-{1}\\{0}&{1}&{2}\\{0}&{0}&-{3}\\{0}&-{1}&-{2}\\{0}&{2}&{1}\end{array}\right]}{\left(\begin{array}{c} {R}_{{2}}\rightarrow{R}_{{2}}+{R}_{{1}}\\{R}_{{3}}\rightarrow{R}_{{3}}-{R}_{{1}}\\{R}_{{4}}\rightarrow{R}_{{4}}-{2}{R}_{{1}}\\{R}_{{5}}\rightarrow{R}_{{5}}-{3}{R}_{{1}}\end{array}\right)}\)
\(\displaystyle\sim{\left[\begin{array}{ccc} {2}&{1}&-{1}\\{0}&{1}&{2}\\{0}&{0}&-{3}\\{0}&{0}&{0}\\{0}&{0}&-{3}\end{array}\right]}{R}_{{4}}\rightarrow{R}_{{4}}+{R}_{{2}},{R}_{{5}}\rightarrow{R}_{{5}}-{2}{R}_{{2}}\)
\(\displaystyle\sim{\left[\begin{array}{ccc} {2}&{1}&-{1}\\{0}&{1}&{2}\\{0}&{0}&-{3}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{array}\right]}{R}_{{5}}\rightarrow{R}_{{5}}-{R}_{{3}}\)
\(\displaystyle{U}={\left[\begin{array}{ccc} {2}&{1}&-{1}\\{0}&{1}&{2}\\{0}&{0}&-{3}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{array}\right]}\)
Step 3
And lower triangular matrix that is L is calculated by the entries are the number in the 0 respectively.
i.e., \(\displaystyle{L}={\left[\begin{array}{ccc} {1}&{0}&{0}\\-{1}&{1}&{0}\\{1}&{0}&{1}\\{2}&-{1}&{0}\\{3}&{2}&{1}\end{array}\right]}\)
0

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