Use python $/$jupyter notebook

Dehghan et al$.(2017)$ developed two$-$step iterative method called the diagonal and off$-$diagonal splitting iteration method for solving systems of linear equations $Ax=B,$ where the coefficient matri:

$A$ is split into its diagonal and off$-$diagonal entries, i$.$e$.,A=D+L+U,$ with

$D=\left[\begin{array}{cccccc}{a}_{11}& 0& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& a_{22}& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& a_{35}& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& 0& a_{44}& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& 0& 0& a_{55}& \text{c}\text{d}\text{o}\text{t}\text{s}\\ \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{d}\text{d}\text{o}\text{t}\text{s}\end{array}\right],L=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ a_{21}& 0& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ a_{31}& a_{32}& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ a_{41}& a_{42}& a_{43}& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ a_{51}& a_{52}& a_{53}& a_{54}& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{d}\text{d}\text{o}\text{t}\text{s}\end{array}\right],U=\left[\begin{array}{cccccc}0& a_{12}& a_{13}& a_{14}& a_{15}& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& a_{23}& a_{24}& a_{25}& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& 0& a_{34}& a_{55}& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& 0& 0& a_{45}& \text{c}\text{d}\text{o}\text{t}\text{s}\\ 0& 0& 0& 0& 0& \text{c}\text{d}\text{o}\text{t}\text{s}\\ \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{v}\text{d}\text{o}\text{t}\text{s}& \text{d}\text{d}\text{o}\text{t}\text{s}\end{array}\right]$

Each iteration involves solving for an intermediate iteration of the solution vector, $x^{(k+\frac{1}{2})},$ from

$D{x}^{(k+\frac{4}{2})}=[{}_{1}D+({}_1-1)L+({}_1-1)U]x^{(k)}+(1-{}_1)B$

that is used to solve for the next iteration of the solution vector, $x^{(k+1)},$ from

$(D+{}_{2}L)x^{(k+1)}=[(1-{}_2)D-{}_{2}U]x^{(k+\frac{1}{2})}+{}_{2}B$

where $0{}_{1}1$ and $0{}_{2}1$ for $k=0,1,2,$dots

The system of equations described by $(1)$ involves a diagonal matrix. Thus, $x^{(k+\frac{1}{2})}$ can be solved efficiently by elementwise division between the constants vector and the main diagonal elements.

The system of equations described by $(2)$ involves a lower triangular matrix Thus, $x^{(k+1)}$ can be solved efficiently via forward subsitution.

You are not allowed to use matrix inversion to solve systems of equations $(1)$ and $(2).$

Part B

Write a Python function $,$ tol $=1E-6)$ for solving a system of linear equations $Ax=B,$ where $A$ is the coefficients matrix, $B$ is the constants vector, $x=$ the

vector of initial values, $w1$ and $w2$ are parameters affecting convergence provided $0w0$ and $0w2=0,$ and tol is the error tolerance. The output of Dos should be $(x,$ err,

$k),$ where $x$ is the solution vector, err is the Euclidean norm of the error vector $x(k+1)-x(k),$ and $k$ corresponds to the number of iterations.

Allow $0$ to accept either a single value or a vector of values. When the input is a single value, your Python function should internally convert it into a vector of initial values, e$.$g$.,0[0,0,$dots,$0]$ or

$1\mathrm{.}23[1\mathrm{.}23,1\mathrm{.}23,$dots,$1\mathrm{.}23].$

Use your function to solve the system of linear equations defined as follows:

$B=(-1{)}^{i}i0im0jnm=n=200x0=0,w1=0\mathrm{.}3,w2=0\mathrm{.}6=0\mathrm{.}02x{x}_{l}ll=1,$dots,$n$