question s: sonving a set on ninear equations

$($In this question, you are supposed to use array structures.$)$

As industrial engineers, we often deal with a set of linear equations. Although there are

exact methods to solve them, they require too much computational effort. Therefore, some

approximation methods have been developed. This question addresses one of them and gives a

small example.

Suppose that you have the following set of linear equations:

$a_{11}{x}_1+a_{12}{x}_2+$dots$+a_{1n}{x}_n=b_1$

$a_{21}{x}_1+a_{22}{x}_2+$dots$+a_{2n}{x}_n=b_2$

$a_{n1}{x}_1+a_{n2}{x}_2+$dots$+a_{}{x}_n=b_n$

In this set $a_{ij}\text{'}$s correspond to the coefficients of the variables $x_j\text{'}$s of the $i^{\text{t}\text{h}\text{e}}$ equation, and

$b_i$ to the right$-$hand side constants, where $i=1,$dots,$n$ and $j=1,$dots,$n.$

The linear equations are rewritten so that the first equation is solved for $x_1,$ the second for

$x_2$ and so on$.$ In other words, the followings are obtained:

$x_1=\frac{b_1-a_{12}{x}_2-\text{d}\text{o}\text{t}\text{s}-a_{1n}{x}_n}{a_{11}}$

$x_2=\frac{b_2-a_{21}{x}_1-\text{d}\text{o}\text{t}\text{s}-a_{2n}{x}_n}{a_{22}}$

vdots

$x_n=\frac{b_n-a_{n1}{x}_1-\text{d}\text{o}\text{t}\text{s}-a_{n,n-1}{x}_{n-1}}{a_{}}$

The method starts with an initial solution, say $x^{(0)}=(x_1^{(0)},x_2^{(0)},$dots,$x_n^{(0)}).$ One may take

any value to start $($note that changing the initial solution only affects the number of iterations

needed to approximate the solution$).$

In each iteration, current value of $x^{(i)}$ is used to calculate the new solution $x^{(i+1)}.$ For

example, $x_1^{(0)},x_2^{(0)},$dots,$x_n^{(0)}$ are substituted in the equations given in $(1),$ and the left$-$hand side

gives the new solution as $x^{(1)}.$ Continuing in this manner, approximate the real solution.

Consider the numerical example below with the application of the iterative approximation

method.

$5x_1-2x_2+3x_3=-1$

$-3x_1+9x_2+x_3=2$

$2x_1-x_2-7x_3=3$

When we rearrange the equations we get,

$x_1=\frac{-1+2x_2-3x_3}{5}$

$x_2=\frac{2+3x_1-x_3}{9}$

$x_3=\frac{-3+2x_1-x_2}{7}$

Taking the initial solution as $x^{(0)}=(0,0,0),$ we get $(x_1^{(1)},x_2^{(1)},x_3^{(1)})=(\frac{-1}{5},\frac{2}{9},\frac{-3}{7})$ at the end

of the first iteration. With the obtained $x^{(1)},$ we continue the procedure. We substitute the

values in the formulations in $(1)$ and get $x^{(2)}$ as $(x_1^{(2)},x_2^{(2)},x_3^{(2)})=(\frac{46}{315},\frac{64}{315},\frac{-163}{315}).$ The method

iterates until the stopping criterion is reached.

Your task is to write a MATLAB script that works on the following set of linear equations:

$9x_1+x_2+x_3+2x_4=12$

$2x_1+10x_2+3x_3-x_4=18$

$3x_1+4x_2+11x_3+x_4=1$

$x_1+2x_2-x_3+x_4=7$

with three different initial solutions:

$x^{(0)}=(x_1^{(0)},x_2^{(0)},x_3^{(0)},x_4^{(0)})=(0,0,0,0),$

$y^{(0)}=(y_1^{(0)},y_2^{(0)},y_3^{(0)},y_4^{(0)})=(10,10,10,10),$

$z^{(0)}=(z_1^{(0)},z_2^{(0)},z_3^{(0)},z_4^{(0)})=(1,1,1,1).$

Take the stopping criterion as the error between the current and previous solution. In other

words, if you are at iteration $k,$ the error is

Error $=\sqrt[\phantom{2}]{(x_1^k-x_1^{k-1}{)}^2+(x_2^k-x_2^{k-1}{)}^2+(x_3^k-x_3^{k-1}{)}^2+(x_4^k-x_4^{k-1}{)}^2}$

Continue until the error is less than the stopping criterion. In this homework, consider three

different stopping criteria as ${10}^{-2},{10}^{-4},$ and ${10}^{-6}.$ Report your results as given in the output

format and comment on your results by putting your explanations as comment lines at the end

of your script.