# Compute the LU factorization of each of the following matrices. begin{bmatrix}1 & 1&1 3 & 5&6-2&2&7 end{bmatrix}

Compute the LU factorization of each of the following matrices.
$\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 6\\ -2& 2& 7\end{array}\right]$
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l1koV
Step 1: to determine
Given:
$A=\left(\begin{array}{ccc}1& 1& 1\\ 3& 5& 6\\ -2& 2& 7\end{array}\right)$ to determine:
LU decomposition of given matrix
Step 2: calculation
Calculation:
$Let\left(\begin{array}{ccc}1& 1& 1\\ 3& 5& 6\\ -2& 2& 7\end{array}\right)=\left(\begin{array}{ccc}{l}_{11}& 0& 0\\ {l}_{21}& {l}_{22}& 0\\ {l}_{31}& {l}_{32}& {l}_{33}\end{array}\right)\left(\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ 0& {u}_{22}& {u}_{23}\\ 0& 0& {u}_{33}\end{array}\right)$
Let
So, $\left(\begin{array}{ccc}1& 1& 1\\ 3& 5& 6\\ -2& 2& 7\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ {l}_{21}& 1& 0\\ {l}_{31}& {l}_{32}& 1\end{array}\right)\left(\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ 0& {u}_{22}& {u}_{23}\\ 0& 0& {u}_{33}\end{array}\right)$
Now multiplying two matrices on RHS , we get,
$\left(\begin{array}{ccc}1& 1& 1\\ 3& 5& 6\\ -2& 2& 7\end{array}\right)=\left(\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ {l}_{21}{u}_{11}& {l}_{21}{u}_{12}+{u}_{22}& {l}_{21}{u}_{13}+{u}_{23}\\ {l}_{31}{u}_{11}& {l}_{31}{u}_{12}+{l}_{32}{u}_{22}& {l}_{31}{u}_{13}+{l}_{32}{u}_{23}+{u}_{33}\end{array}\right)$
Step 3
Now , equating each entry , we get
${u}_{11}=1$
${u}_{12}=1$
${u}_{13}=1$
${l}_{21}{u}_{11}=3⇒{l}_{21}\cdot 1=3⇒{l}_{21}=3$
${l}_{21}{u}_{12}+{u}_{22}=5⇒3.1+{u}_{22}=5⇒{u}_{22}=2$
${l}_{21}{u}_{13}+{u}_{23}=6⇒3.1+{u}_{23}=6⇒{u}_{23}=3$
${l}_{31}{u}_{11}=-2⇒{l}_{31}\cdot 1=-2⇒{l}_{31}=-2$
${l}_{31}{u}_{12}+{l}_{32}{u}_{22}=2⇒-2.1+{l}_{32}\cdot 2=2⇒{l}_{32}=2$
${l}_{31}{u}_{13}+{l}_{32}{u}_{23}+{u}_{33}=7⇒-2.1+2.3+{u}_{33}=7⇒{u}_{33}=3$
So,
Jeffrey Jordon