# Even though it is inefficient, can a linear system be solved using Newton-Raphson?

Given Simultaneous linear equations of the form, $$a_{11}x_1 + a_{12}x_2+a_{13}x_3+\cdots a_{1n}x_n = b_1$$ $$a_{21}x_1 + a_{22}x_2+a_{23}x_3+\cdots a_{2n}x_n=b_2$$ $$a_{31}x_1 + a_{32}x_2+a_{33}x_3+\cdots a_{3n}x_n=b_3$$ $$\vdots$$ $$a_{n1}x_1 + a_{n2}x_2+a_{n3}x_3+\cdots a_{nn}x_n=b_n$$

Can the Newton Raphson's Method use to solve this system of linear equations? Please mention the reasons and possible arguments.

• In the case of a linear function, what is the difference? – Matthew Leingang Sep 10 '19 at 11:56
• @MatthewLeingang In the Case of Linear functions, the inverse of Jacobian is a constant matrix. – Akash Tadwai Sep 10 '19 at 12:00
• Are you sure? What's the Jacobian of a linear function? – Matthew Leingang Sep 10 '19 at 12:02
• Applying the Newton-Raphson method involves calculating the inverse of a function. In this case, it is equivalent to inverting the coefficient matrix. So basically it's the same thing as just solving the system the normal way. – Matti P. Sep 10 '19 at 12:04

• I know that there are relaxation methods, but I'm just asking that is the process is same as relaxation methods when we are solving a system of linear equations? $x^{k+1}=x^k-[{A^{(k)}}]^{-1} F[x^{(k)}]$, As inverse is a constant matrix does it mean the same? – Akash Tadwai Sep 10 '19 at 12:13
• @AkashTadwai Yes. If you want to find a root of $f(x)=Ax-b$, the Jacobian is $J(x)=A$ so the Newton step converges in one step for any initial guess of $x$. – Algebraic Pavel Sep 10 '19 at 13:02