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$2 \times 2$ matrix
The method outlined so far will be applied to the simplest non-trivial example,
which is a matrix of rank $2$. The system of equations to be solved is:
$$
\left[ \begin{array}{cc} a & b \\
c & d \end{array} \right]
\left[ \begin{array}{c} x \\
y \end{array} \right] =
\left[ \begin{array}{c} p \\
q \end{array} \right]
$$
Assume that $a \neq 0$ and $d \neq 0$. At first, the equations will be normed:
$$
\left[ \begin{array}{cc} 1 & b/a \\
c/d & 1 \end{array} \right]
\left[ \begin{array}{c} x \\
y \end{array} \right] =
\left[ \begin{array}{c} p/a \\
q/d \end{array} \right]
$$
The matrix $M$ is formed by subtracting from the identity matrix:
$$
M = \left[ \begin{array}{cc} 0 & -b/a \\
-c/d & 0 \end{array} \right]
$$
Hence:
$$
I + M = \left[ \begin{array}{cc} 1 & -b/a \\
-c/d & 1 \end{array} \right]
$$
And:
$$
M^2 = \left[ \begin{array}{cc} 0 & -b/a \\
-c/d & 0 \end{array} \right] .
\left[ \begin{array}{cc} 0 & -b/a \\
-c/d & 0 \end{array} \right] =
\left[ \begin{array}{cc} b/a.c/d & 0 \\
0 & c/d.b/a \end{array} \right]
$$
Hence:
$$
I - M^2 = \left[ \begin{array}{cc} 1 - b/a.c/d & 0 \\
0 & 1 - c/d.b/a \end{array} \right]
$$
Nothing is simpler than determining the inverse of a diagonal matrix:
$$
(I - M^2)^{-1} = \left[ \begin{array}{cc} 1/(1 - b/a.c/d) & 0 \\
0 & 1/(1 - c/d.b/a) \end{array} \right]
$$
Hence:
$$
\frac{I + M}{I - M^2} =
\left[ \begin{array}{cc} 1/(1 - b/a.c/d) & 0 \\
0 & 1/(1 - c/d.b/a) \end{array} \right] .
\left[ \begin{array}{cc} 1 & -b/a \\
-c/d & 1 \end{array} \right] =
$$ $$
\frac{1}{1 - b/a.c/d} .
\left[ \begin{array}{cc} 1 & -b/a \\
-c/d & 1 \end{array} \right]
$$
It is easily recognized that this is exactly the inverse of:
$$
\left[ \begin{array}{cc} 1 & b/a \\
c/d & 1 \end{array} \right]
$$
It is concluded that any system of $2$ equations with $2$ unknowns is solved
exactly by the Newton-Raphson method (provided that the main diagonal
elements are non-zero and the matrix is non-singular).