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
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

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}$$

Relevant Questions

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Determine whether the given set S is a subspace of the vector space V.
A. V=$$P_5$$, and S is the subset of $$P_5$$ consisting of those polynomials satisfying p(1)>p(0).
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C. $$V=R^n$$, and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.
D. V=$$C^2(I)$$, and S is the subset of V consisting of those functions satisfying the differential equation y″−4y′+3y=0.
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Compute the LU factorization of each of the following matrices.
$$\begin{bmatrix}-2 & 1 & 2 \\4 & 1 & -2 \\-6 & -3 & 4\end{bmatrix}$$
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when $$Z=1.3$$ and $$H=0.05$$;
Assume that you do not have vales of the area beyond $$z=1.2$$ in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table $$[for\ z=1.3\ and\ H=0.05]$$.
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$
Solve the system $$Ax = b$$ using the given LU factorization of A
$$A=\begin{bmatrix}2 & -4 & 0 \\3 & -1 & 4 \\ -1 & 2 & 2\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\\frac{3}{2} & 1 & 0 \\ -\frac{1}{2} & 0 & 1\end{bmatrix}\times \begin{bmatrix}2 & -4 & 0 \\0 & 5 & 4 \\ 0 & 0 & 2\end{bmatrix}, b=\begin{bmatrix}2 \\0 \\-5\end{bmatrix}$$
$$A=\begin{bmatrix}2 & 1 & -1 \\-2 & 0 & 3 & \\ 2 & 1 & -4\\4 & 1 & -4 \\ 6 & 5 & -2\end{bmatrix}$$