Let a linear sytem of equations Ax=b where A=begin{pmatrix}4 & 2&-2 2 & 2&-3-2&-3&14 end{pmatrix} , b=begin{pmatrix}10 , 5 , 4 end{pmatrix}^T in case we solve this equation system by using Dolittle LU factorization method , find Z and X matrices

Let a linear sytem of equations Ax=b where
$A=\left(\begin{array}{ccc}4& 2& -2\\ 2& 2& -3\\ -2& -3& 14\end{array}\right),b={\left(\begin{array}{c}10,5,4\end{array}\right)}^{T}$
in case we solve this equation system by using Dolittle LU factorization method , find Z and X matrices
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Step 1
By using Dolitte LU decomposition method ,
A=LU
$\left[\begin{array}{ccc}4& 2& -2\\ 2& 2& -3\\ -2& -3& 14\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ {L}_{21}& 1& 0\\ {L}_{31}& {L}_{32}& 1\end{array}\right]\left[\begin{array}{ccc}{u}_{11}& {u}_{12}& {u}_{13}\\ 0& {u}_{22}& {u}_{23}\\ 0& 0& {u}_{33}\end{array}\right]$
$\left[\begin{array}{ccc}4& 2& -2\\ 2& 2& -3\\ -2& -3& 14\end{array}\right]=\left[\begin{array}{ccc}{u}_{11}& {u}_{11}& {u}_{11}\\ {u}_{11}{L}_{21}& {L}_{21}{u}_{12}+{u}_{22}& {L}_{21}{u}_{13}+{u}_{23}\\ {L}_{31}{u}_{11}& {u}_{12}{L}_{31}+{L}_{32}{u}_{22}& {L}_{31}{u}_{13}+{L}_{32}{u}_{23}+{u}_{33}\end{array}\right]$
Step 2
By comparing the two matrices we will get,
${u}_{11}=4$
${u}_{12}=2$
${u}_{13}=-2$
${u}_{11}{L}_{21}=2$
$4{L}_{21}=2$
${L}_{21}=0.5$
${L}_{21}{u}_{12}+{u}_{22}=2$
${u}_{22}=2$
${L}_{21}{u}_{13}+{u}_{23}=-3$
${u}_{23}=-2$
${L}_{31}{u}_{11}=-2$
${L}_{31}=-0.5$
${u}_{12}{L}_{31}+{L}_{32}{u}_{22}=-3$
${L}_{32}=-0.5$
${L}_{31}{u}_{13}+{L}_{32}{u}_{23}+{u}_{33}=14$
${u}_{33}=9$
Step 3
Therefore, the matrices L and U will be ,
$L=\left[\begin{array}{ccc}1& 0& 0\\ 0.5& 1& 0\\ -0.5& -2& 1\end{array}\right]$
$U=\left[\begin{array}{ccc}4& 2& -2\\ 0& 1& -2\\ 0& 0& 9\end{array}\right]$ Let Lz=b where $z=\left[\begin{array}{c}{z}_{1}\\ {z}_{2}\\ {z}_{3}\end{array}\right]$
Therefore, we will get ,
$\left[\begin{array}{ccc}1& 0& 0\\ 0.5& 1& 0\\ -0.5& -2& 1\end{array}\right]\left[\begin{array}{c}{z}_{1}\\ {z}_{2}\\ {z}_{3}\end{array}\right]=\left[\begin{array}{c}10\\ 5\\ 4\end{array}\right]$
Step 4
From these we get,
${z}_{1}=10$
$0.5{z}_{1}+{z}_{2}=5$
${z}_{2}=0$
$-\frac{1}{2}{z}_{1}-2{z}_{2}+{z}_{3}=4$
${z}_{3}=9$
Thus t

Jeffrey Jordon