Find an LU factorization of the matrix A (with L unit lower triangular). A=[(5,4),(-4,-3)] L-? U-?

Find an LU factorization of the matrix A (with L unit lower triangular). A=[(5,4),(-4,-3)] L-? U-?

Question
Polynomial factorization
asked 2021-02-10
Find an LU factorization of the matrix A (with L unit lower triangular).
\(\displaystyle{A}={\left[\begin{array}{cc} {5}&{4}\\-{4}&-{3}\end{array}\right]}\)
L-?
U-?

Answers (1)

2021-02-11
Step 1
Given that,
\(\displaystyle{A}={\left[\begin{array}{cc} {5}&{4}\\-{4}&-{3}\end{array}\right]}\)
A=LU
\(\displaystyle\Rightarrow{A}={L}{U}={\left[\begin{array}{cc} {1}&{0}\\{L}_{{{21}}}&{1}\end{array}\right]}{\left[\begin{array}{cc} {U}_{{{11}}}&{U}_{{{12}}}\\{0}&{U}_{{{22}}}\end{array}\right]}={\left[\begin{array}{cc} {U}_{{{11}}}&{U}_{{{12}}}\\{L}_{{{21}}}{U}_{{{11}}}&{L}_{{{21}}}{U}_{{{12}}}+{U}_{{{22}}}\end{array}\right]}\)
\(\displaystyle\Rightarrow{\left[\begin{array}{cc} {5}&{4}\\-{4}&-{3}\end{array}\right]}={\left[\begin{array}{cc} {U}_{{{11}}}&{U}_{{{12}}}\\{L}_{{{21}}}{U}_{{{11}}}&{L}_{{{21}}}{U}_{{{12}}}+{U}_{{{22}}}\end{array}\right]}\)
Step 2
By comparing the matrices,
\(\displaystyle\Rightarrow{U}_{{{11}}}={5},{U}_{{{12}}}={4}\)
\(\displaystyle{L}_{{{21}}}{U}_{{{11}}}=-{4}\Rightarrow{L}_{{{21}}}{\left({5}\right)}=-{4}\Rightarrow{L}_{{{21}}}=-\frac{{4}}{{5}}\)
\(\displaystyle{L}_{{{21}}}{U}_{{{12}}}+{U}_{{{22}}}=-{3}\Rightarrow-\frac{{4}}{{5}}{\left({4}\right)}+{U}_{{{22}}}=-{3}\Rightarrow{U}_{{{22}}}=-{3}+\frac{{16}}{{5}}\Rightarrow{U}_{{{22}}}=\frac{{1}}{{5}}\)
We have,
\(\displaystyle{L}={\left[\begin{array}{cc} {1}&{0}\\{L}_{{{21}}}&{1}\end{array}\right]},{U}={\left[\begin{array}{cc} {U}_{{{11}}}&{U}_{{{12}}}\\{0}&{U}_{{{22}}}\end{array}\right]}\)
Plug the know values in matrices, get
\(\displaystyle{L}={\left[\begin{array}{cc} {1}&{0}\\-\frac{{4}}{{5}}&{1}\end{array}\right]},{U}={\left[\begin{array}{cc} {5}&{4}\\{0}&\frac{{1}}{{5}}\end{array}\right]}\)
0

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