Create a $44$ matrix A with ones on and above the diagonal. Also set vector$-$column $x_0=([1\mathrm{.}012\mathrm{.}01,3\mathrm{.}014\mathrm{.}01]).$ Assuming this $x_0$ is the solution to the linear system $A{x}_0=b_0,$ find the corresponding right$-$hand side $b_0.$ Report to hw$03$q$1.$m$.$ Hint: use $A**x_{-}0$ for the matrix$-$vector product $A{x}_0$ of the matrix A and the vector $x_0.$

Create a column right$-$hand side $7\mathrm{.}034\mathrm{.}03).$ Show that $x_0$ above does not solve the linear system $Ax=b$ by computing the residual $r=b-A{x}_0$ and comparing it with the zero vector. Add to hw$03$q$1.$m$.$

Solve the system $Ax=b$ on paper using exact arithmetic.

Using BackwardSubstitution.m$,$ verify your computation on paper above. Add to hw$03$q$1.$m$.$

Since $b=Ax$ and $b_0=A{x}_0,$ notice that it must hold that

$A(x-x_0)=b-b_0$

meaning the mantra "a perturbation in the right$-$hand side causes a perturbation in the solution up to the multiplication by $A".$ Verify the latter $($up to machine epsilon$)$ and report it to hw$03$q$1.$m