Oberlaudacu

2021-12-28

Find the LU-factorization of following matrix.
$A=\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]$

trisanualb6

Step 1
$A=\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]$
LU factorization of A
$A=LU$
Where L is lower triangular matrix.
U is upper triangular matrix.
We obtain upper mangular matrix U by Gaussian Elimination method and L is made upd multiplies we used in Gaussian Elimination with 1 at diagonal enties.
Step 2
$A=\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]{R}_{2}-3{R}_{1}⇒{L}_{2,1}=3\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 2& 6& 13\end{array}\right]$
$\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 2& 6& 13\end{array}\right]{R}_{3}-2{R}_{1}⇒{L}_{3,1}=2\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 0& 2& 5\end{array}\right]{R}_{3}-{R}_{2}⇒{L}_{3,2}=1\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$
$\therefore U=\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$
and $L=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 1& 1\end{array}\right]\left[\begin{array}{c}{L}_{2,1}=3\\ {L}_{3,1}=2\\ {L}_{3,2}=1\end{array}\right]$
$A=LU=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 1& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$

Barbara Meeker

$A=\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]$
and $L=\left[\begin{array}{ccc}1& 0& 0\\ {l}_{21}& 1& 0\\ {l}_{31}& {l}_{32}& 1\end{array}\right]$
and $KU=\left[\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ 0& {u}_{22}& {u}_{23}\\ 0& 0& {u}_{33}\end{array}\right]$
Now $A=LU$
$\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ {l}_{21}& 1& 0\\ {l}_{31}& {l}_{32}& 1\end{array}\right]\left[\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ 0& {u}_{22}& {u}_{23}\\ 0& 0& {u}_{33}\end{array}\right]$
$\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]=\left[\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ {l}_{21}{u}_{11}& {l}_{21}{u}_{22}+{u}_{22}& {l}_{22}{u}_{23}+{u}_{23}\\ {l}_{31}{u}_{11}& {l}_{31}{u}_{12}+{l}_{32}{u}_{22}& {l}_{31}{u}_{13}+{l}_{32}{u}_{23}+{u}_{33}\end{array}\right]$
Now, compaing both the sides, then
${u}_{11}=1,{u}_{12}=2,{u}_{13}=4$
${l}_{21}{u}_{11}=3⇒{l}_{21}×1=3⇒{l}_{21}=3$
${l}_{21}{u}_{12}+{u}_{22}=8⇒3×2+{u}_{22}=8$
${u}_{22}=2$
${l}_{21}{u}_{13}+{u}_{23}=14⇒3×4+{u}_{23}=14$
${u}_{23}=14-12=2$
${u}_{23}=2$
${l}_{31}{u}_{11}=2⇒{l}_{31}×1=2⇒{l}_{31}=2$
${l}_{31}{u}_{12}+{l}_{32}{u}_{22}=6$

karton

Let

$\begin{array}{}A=\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]\\ A=LU\\ L=\left[\begin{array}{ccc}1& 0& 0\\ {L}_{21}& 1& 0\\ {L}_{31}& {L}_{32}& 1\end{array}\right],U=\left[\begin{array}{ccc}{U}_{11}& {U}_{12}& {U}_{13}\\ 0& {U}_{22}& {U}_{23}\\ 0& 0& {U}_{33}\end{array}\right]\\ \left[\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ {l}_{21}{u}_{11}& {l}_{21}{u}_{12}+{u}_{22}& {l}_{21}{u}_{13}+{u}_{23}\\ {l}_{31}{u}_{11}& {l}_{31}{u}_{12}+{l}_{32}{u}_{22}& {l}_{31}{u}_{13}+{l}_{32}{u}_{23}+{u}_{33}\end{array}\right]\\ =\left[\begin{array}{ccc}1& 2& 4\\ 3& 8& 14\\ 2& 6& 13\end{array}\right]\\ {u}_{11}=1\\ {u}_{12}=2\\ {u}_{13}=4\\ {l}_{21}{u}_{11}=3,{l}_{21}{x}_{1}=3⇒{l}_{21}=3\\ {l}_{21}{u}_{12}+{u}_{22}=8\therefore 3×2+{u}_{22}=8\\ {u}_{22}=2\\ {l}_{21}{u}_{13}+{u}_{23}=14⇒3×4+{u}_{23}=14\\ {u}_{23}=2\\ {l}_{31}{u}_{11}=2⇒{l}_{31}=2\\ {l}_{31}{u}_{12}+{l}_{32}{u}_{32}=6⇒2×2+{l}_{32}×2=6\\ {l}_{32}=1\\ {l}_{31}{u}_{13}+{l}_{32}{u}_{23}+{u}_{33}=13⇒8+2+{u}_{33}=13\\ {u}_{33}=3\\ A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 1& 1\end{array}\right]=L,\left[\begin{array}{ccc}1& 2& 4\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]=U\end{array}$
This is the LU decomposition.

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