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

Question
Polynomial factorization
find an LU factorization of the given matrix.
$$\displaystyle{\left[\begin{array}{ccc} {2}&{2}&-{1}\\{4}&{0}&{4}\\{3}&{4}&{4}\end{array}\right]}$$

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
$$\displaystyle{L}={\left[\begin{array}{ccc} {1}&{1}&{0}\\{2}&{1}&{0}\\\frac{{3}}{{2}}&-\frac{{1}}{{4}}&{1}\end{array}\right]}$$

### Relevant Questions

Find the LU-Factorization of the matrix A below
$$\displaystyle{A}={\left[\begin{array}{ccc} {2}&{1}&-{1}\\-{2}&{0}&{3}\\{2}&{1}&-{4}\\{4}&{1}&-{4}\\{6}&{5}&-{2}\end{array}\right]}$$
Find an LU factorization of the matrix A (with L unit lower triangular).
$$\displaystyle{A}={\left[\begin{array}{ccc} -{4}&{0}&{4}\\{12}&{2}&-{9}\\{12}&{8}&{9}\end{array}\right]}$$
L-?
U-?
A given polynomial has a root of x = 3, an zero of x = -2, and an x-intercept of x = -1.
The equation of this polynomial in factored form would be f(x) =
The equation of this polynomial in standard form would be f(x) =
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-?
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
How can an inverse of a modulo m be used to solve the congruence $$\displaystyle{a}{x}\equiv{b}{\left({b}\text{mod}{m}\right)}$$ when gcd(a,m)=1.
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{23}{x}^{{{3}}}-{13}{x}^{{{2}}}+{32}{x}+{16}$$
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}$$
$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{14}{x}^{{{4}}}-{14}{x}^{{{3}}}+{36}{x}^{{{2}}}+{43}{x}+{10}$$
To solve: The system of congruence $$\displaystyle{x}={2}{\left({b}\text{mod}{3}\right)},{x}={1}{\left({b}\text{mod}{4}\right)},\ {\quad\text{and}\quad}\ {x}={3}{\left({b}\text{mod}{5}\right)}$$ using the method of back substitution.