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=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

Question
Matrices
asked 2021-03-12
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

Answers (1)

2021-03-13
Step 1
By using Dolitte LU decomposition method ,
A=LU
\(\begin{bmatrix}4 & 2&-2 \\2 & 2&-3\\-2&-3&14 \end{bmatrix}=\begin{bmatrix}1 & 0&0 \\L_{21} & 1&0\\L_{31}&L_{32}&1 \end{bmatrix}\begin{bmatrix}u_{11} & u_{12}&u_{13} \\0 & u_{22}&u_{23}\\0&0&u_{33} \end{bmatrix}\)
\(\begin{bmatrix}4 & 2&-2 \\2 & 2&-3\\-2&-3&14 \end{bmatrix}=\begin{bmatrix}u_{11} & u_{11}&u_{11} \\u_{11}L_{21} & L_{21}u_{12}+u_{22}&L_{21}u_{13}+u_{23}\\L_{31}u_{11}&u_{12}L_{31}+L_{32}u_{22}&L_{31}u_{13}+L_{32}u_{23}+u_{33} \end{bmatrix}\)
Step 2
By comparing the two matrices we will get,
\(u_{11}=4\)
\(u_{12}=2\)
\(u_{13}=-2\)
\(u_{11}L_{21}=2\)
\(4L_{21}=2\)
\(L_{21}=0.5\)
\(L_{21}u_{12}+u_{22}=2\)
\(u_{22}=2\)
\(L_{21}u_{13}+u_{23}=-3\)
\(u_{23}=-2\)
\(L_{31}u_{11}=-2\)
\(L_{31}=-0.5\)
\(u_{12}L_{31}+L_{32}u_{22}=-3\)
\(L_{32}=-0.5\)
\(L_{31}u_{13}+L_{32}u_{23}+u_{33}=14\)
\(u_{33}=9\)
Step 3
Therefore, the matrices L and U will be ,
\(L=\begin{bmatrix}1 & 0&0 \\0.5 & 1&0\\-0.5&-2&1 \end{bmatrix}\)
\(U=\begin{bmatrix}4 & 2&-2 \\0 & 1&-2\\0&0&9 \end{bmatrix} Let Lz=b where \(z=\begin{bmatrix}z_1\\z_2\\z_3 \end{bmatrix}\)
Therefore, we will get ,
\(\begin{bmatrix}1 & 0&0 \\0.5 & 1&0\\-0.5&-2&1 \end{bmatrix}\begin{bmatrix}z_1\\z_2\\z_3 \end{bmatrix}=\begin{bmatrix}10\\5\\4 \end{bmatrix}\)
Step 4
From these we get,
\(z_1=10\)
\(0.5z_1+z_2=5\)
\(z_2=0\)
\(-\frac{1}{2}z_1-2z_2+z_3=4\)
\(z_3=9\)
Thus the matrix Z is ,
\(Z=\begin{bmatrix}10\\0\\9 \end{bmatrix}\)
Step 5
The matrix equation will be ,
\(\begin{bmatrix}4 & 2&-2 \\0&1&-2\\0&0&9 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}10 \\0\\9 \end{bmatrix}\)
The equations will be,
\(9x_3=9\)
\(x_3=1\)
\(x_2-2x_3=0\)
\(x_2=2\)
\(4x_1+2x_2-2x_3=10\)
\(x_1=2\)
Step 6
The matrix of X will be,
\(X=\begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix}\)
\(=\begin{bmatrix}2 \\2\\1 \end{bmatrix}\)
Step 7
CONCLUSION :
The values of Z and X are as follows :
\(Z=\begin{bmatrix}10 \\0\\9 \end{bmatrix}\)
\(X=\begin{bmatrix}2 \\2\\1 \end{bmatrix}\)
0

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