find an LU factorization of the given matrix.
[(2,2,-1),(4,0,4),(3,4,4)]

find an LU factorization of the given matrix.
[(2,2,-1),(4,0,4),(3,4,4)]

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Answers (1)

2021-03-09

Step 1
To solve, we use the multiplier method.
In an elementary row operation is of the form \(\displaystyle{R}_{{i}}-{k}{R}_{{j}}\) , where k is called the scalar multiplier.
\(\displaystyle{A}={\left[\begin{array}{ccc} {2}&{2}&-{1}\\{4}&{0}&{4}\\{3}&{4}&{4}\end{array}\right]}\) \(\displaystyle={\left[\begin{array}{ccc} {2}&{2}&-{1}\\{0}&-{4}&{6}\\{0}&{1}&\frac{{11}}{{2}}\end{array}\right]}{R}_{{2}}\rightarrow{R}_{{2}}-{2}{R}_{{1}}\) \(\displaystyle{R}_{{3}}\rightarrow{R}_{{3}}-\frac{{3}}{{2}}{R}_{{1}}\) \(\displaystyle={\left[\begin{array}{ccc} {2}&{2}&-{1}\\{0}&-{4}&{6}\\{0}&{0}&{7}\end{array}\right]}{R}_{{3}}\rightarrow{R}_{{3}}-{\left(-\frac{{1}}{{4}}\right)}{R}_{{2}}\)
Step 2
The multipliers were \(\displaystyle{I}_{{{21}}}={2},{I}_{{{31}}}=\frac{{3}}{{2}}{\quad\text{and}\quad}{I}_{{{32}}}=-\frac{{1}}{{4}}\) , so we get
\(\displaystyle{L}={\left[\begin{array}{ccc} {1}&{1}&{0}\\{2}&{1}&{0}\\\frac{{3}}{{2}}&-\frac{{1}}{{4}}&{1}\end{array}\right]}\)

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