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