Methods for Solving Linear Algebraic Equations: Which One Is Best?

[eddysd]

For the system of linear algebraic equations:
x − y = 1
2x + 3y = 7
a) find solution by the Cramer’s rule;
b) find solution by the method of Gaussian elimination;
c) write the first iteration of the Jacobi method with the initial guess x^0=y^0=1.

I understand Cramer's Rule, and I think I understand Gaussian elimination, have got
x=2 and y=1 for both of them, but I have no idea how to do part c). Can anyone help me?

 

[HallsofIvy]

In terms of matrices, this system would be written as
\begin{bmatrix}1 & -1 \\ 2 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ 7\end{bmatrix}

We can take out the "diagonal" and rewrite it as
\begin{bmatrix}1 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}i+ \begin{bmatrix}0 & -1 \\2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}1 \\ 7\end{bmatrix}

so that
\begin{bmatrix}1 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}1 \\ 7\end{bmatrix}- \begin{bmatrix}0 & -1 \\2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}

or, just in terms of the equations, x= 1+ y, 3y= 7- 2x so we have x= 1+ y, y= (7- 2x)/3.

Taking x^0= y^0= 1, we have x^1= 1+ y^0= 1+ 1= 2 and y^1= (7- 2x^0)/3= (7- 2)/3= 5/3.

 

[eddysd]

Thank you, very helpful!