Gaussian Elimination with Scaled Row Pivoting for numerical methods I am solving a system first with basic Gaussian Elimination, and then Gaussian Elimination with scaled row pivoting (used in numerical methods)
Basic Gaussian Elimination on the system $Ax=b$:
\begin{equation}                                                                                                                                                            
\begin{pmatrix}-1& 1& -4 \\                                                                                                                                                 
                2& 2& 0 \\                                                                                                                                                  
                3& 3& 2 \end{pmatrix}                                                                                                                                       
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =                                                                                                                                 
\begin{pmatrix}0\\1\\\frac{1}{2}\end{pmatrix}                                                                                                                               
\end{equation}
Let $A_i$ denote the $i^{th}$ row of matrix $A$ and let $A^{(1)} A^{(2)}...$ denote the matrix after the first, second and so forth elementary row operations. Note that
$A^{0} =A$.
Compute the following elementary row operations:
\begin{align}                                                                                                                                                               
A^{(1)}_2 =& A^{(0)}_2 - (-2)A^{(0)}_1 \\                                                                                                                                   
A^{(1)}_3 =& A^{(0)}_3 - (-3)A^{(0)}_1                                                                                                                                       
\end{align}
This yields:
This yields:
\begin{equation}                                                                                                                                                                                                                                                                                             
\begin{pmatrix}-1& 1& -4 \\                                                                                                                                                                                                                                                                                  
                0& 4& -8 \\                                                                                                                                                                                                                                                                                  
                0& 6& -10 \end{pmatrix}                                                                                                                                                                                                                                                                      
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =                                                                                                                                                                                                                                                                  
\begin{pmatrix}0\\1\\-1\end{pmatrix}                                                                                                                                                                                                                                                                         
\end{equation}
Compute:
\begin{equation} A^{(2)}_3 = A^{(1)}_3 - (\frac{3}{2})A^{(1)}_2\end{equation}
This yields:
\begin{equation}                                                                                                                                                                                                                                                                                             
\begin{pmatrix}-1& 1& -4\\                                                                                                                                                                                                                                                                                   
                0& 4& -8\\                                                                                                                                                                                                                                                                                   
                0& 0& 2\end{pmatrix}                                                                                                                                                                                                                                                                         
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=                                                                                                                                                                                                                                                                   
\begin{pmatrix}0\\1\\-1\end{pmatrix}                                                                                                                                                                                                                                                                         
\end{equation}
Thus we have:
\begin{equation} x=\begin{pmatrix}\frac{5}{4}\\                                                                                                                                                                                                                                                              
                                \frac{-3}{4}\\                                                                                                                                                                                                                                                               
                                \frac{-1}{2}\end{pmatrix}\end{equation}
Now I will solve the same system with Scaled Row Pivoting. The $i^{th}$ element of the list $S$ will denote the maximum element in row $i$ in matrix $A$. $P$ will denote the order of the rows. Initially we have:
\begin{equation}                                                                                                                                                                                                                                                                                             
S = (4, 2, 3) \\                                                                                                                                                                                                                                                                                             
P = (2, 1, 3)                                                                                                                                                                                                                                                                                                
\end{equation}
Swap rows $1$ and $2$ since row $2$ has the maximum pivot relative to its row:
\begin{equation}                                                                                                                                                                                                                                                                                             
\begin{pmatrix}2&2&0\\                                                                                                                                                                                                                                                                                       
                -1&1&-4\\                                                                                                                                                                                                                                                                                    
                3& 3& 2\end{pmatrix}                                                                                                                                                                                                                                                                         
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =                                                                                                                                                                                                                                                                  
\begin{pmatrix}1\\0\\\frac{1}{2}\end{pmatrix}                                                                                                                                                                                                                                                                
\end{equation}
Now compute the following elementary row operations w.r.t the ordering given by $p$:
\begin{align}                                                                                                                                                                                                                                                                                                
  A^{(1)}_1 =& A^{(0)}_1 - (\frac{-1}{2})A^{(0)}_2 \\                                                                                                                                                                                                                                                        
  A^{(1)}_3 =& A^{(0)}_3 - (\frac{3}{2})A^{(0)}_2                                                                                                                                                                                                                                                            
\end{align}
This yields:
\begin{equation}                                                                                                                                                                                                                                                                                             
\begin{pmatrix}2&2&0\\                                                                                                                                                                                                                                                                                       
               0&2&-4\\                                                                                                                                                                                                                                                                                      
               0&0&2\end{pmatrix}                                                                                                                                                                                                                                                                            
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=                                                                                                                                                                                                                                                                   
\begin{pmatrix}1\\\frac{1}{2}\\-1                                                                                                                                                                                                                                                                            
\end{pmatrix}                                                                                                                                                                                                                                                                                                
\end{equation}
Now using back substitution to solve for $x$ we get:
\begin{equation}                                                                                                                                                                                                                                                                                             
x=\begin{pmatrix}\frac{-1}{4}\\\frac{3}{4}\\\frac{-1}{2}\end{pmatrix}                                                                                                                                                                                                                                        
\end{equation}
Clearly, I must have made a mistake along the way since the solutions for both methods are not the same! I know that the scaled pivoting is incorrect as I checked my solution in a CAS and it matched the solution for the Basic Method.
Please show me what I have done wrong in the scaled pivoting algorithm.

You miscomputed $x_2$ in the back substitution of the row-pivoted system, that's the origin of the discrepancy.
