Exercise $1$This exercise involves implementing the elimination method to convert a matrix into row$-$echelon form. As discussed in lectures, the primary approach involves inspecting the values along the diagonal. If they equate to $0,$ an attempt to swap rows should be made to obtain a non$-$zero value.# import numpy as npdef row$\_$echelon$\_$form$($A$,$ B$)$: $"""$ Utilizes elementary row operations to transform a given set of matrices, which represent the coefficients and constant terms of a linear system, into row echelon form. Parameters: $-$ A $($numpy$.$array$)$: The input square matrix of coefficients. $-$ B $($numpy$.$array$)$: The input column matrix of constant terms. Returns: numpy.array: A new augmented matrix in row echelon form with pivots as $1."""$ # Before any computation, check if matrix A $($coefficient matrix$)$ has non$-$zero determinant. # It will be used the numpy sub library np$.$linalg to compute it$.$ det$\_$A $=$ np$.$linalg.det$($A$)$ # Returns "Singular system" if determinant is zero if np$.$isclose$($det$\_$A$,0)==$ True: return 'Singular system' # Make copies of the input matrices to avoid modifying the originals A $=$ A$.$copy$()$ B $=$ B$.$copy$()$ # Convert matrices to float to prevent integer division A $=$ A$.$astype$(\text{'}$float$64\text{'})$ B $=$ B$.$astype$(\text{'}$float$64\text{'})$ # Number of rows in the coefficient matrix num$\_$rows $=$ len$($A$)$ # Create the augmented matrix M $($concatenate A and B$)$ M $=$ np$.$hstack$(($A$,$ B$.$reshape$(-1,1)))$ # Iterate over the rows. for row in range$($num$\_$rows$)$: # Get the pivot candidate from the diagonal element pivot$\_$candidate $=$ M$[$row$,$ row$]$ # If pivot$\_$candidate is zero, try to find a non$-$zero pivot below the current row if np$.$isclose$($pivot$\_$candidate, $0)$: for i in range$($row $+1,$ num$\_$rows$)$: if not np$.$isclose$($M$[$i$,$ row$],0)$: # Swap rows M$[[$row$,$ i$]]=$ M$[[$i$,$ row$]]$ pivot$\_$candidate $=$ M$[$row$,$ row$]$ # Recalculate the pivot after swap break # If pivot$\_$candidate is non$-$zero, scale the row to make the pivot equal to $1$ if not np$.$isclose$($pivot$\_$candidate, $0)$: M$[$row$]=$ M$[$row$]/$ pivot$\_$candidate # Perform row reduction for rows below the current row for i in range$($row $+1,$ num$\_$rows$)$: # Get the factor to eliminate the value below the pivot factor $=$ M$[$i$,$ row$]$ # Perform row reduction: row$\_$to$\_$reduce $->$ row$\_$to$\_$reduce $-$ factor $*$ pivot$\_$row M$[$i$]=$ M$[$i$]-$ factor $*$ M$[$row$]$ return M