PYTHON CODE MODIFICATION The below code is used for solving a system of linear equations using Gaus-Jordan Elimination: # Define the matrix A and vector b A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]] b = [[1], [-1], [2]] # Combine A and b into augmented matrix C C = [row_a + row_b for row_a, row_b in zip(A, b)] # Get the number of rows and columns of the matrix n = len(C) # Set E to 1 (initially assume a unique solution exists) E=1 # Iterate through columns of matrix for j in range(n): # Find the row with the largest magnitude value in column j, only looking at rows j and below pivot_row = max(range(j, n), key=lambda i: abs(C[i][j])) p = C[pivot_row][j] # Check if the pivot is zero if p == 0: E=0 break # If the pivot is not in row j, swap rows j and the pivot row if pivot_row !=j: C[j], C[pivot_row] = C[pivot_row], C[j] # Divide row j by pivot value C[j] = [C[j][i] / p for i in range(n)] # Subtract the pivot row from all other rows to eliminate the value at column j for i in range(n): if i != j: factor = C[i][j] C[i] = [C[i][k] - factor * C[j][k] for k in range(n)] # Check if a unique solution was found if E == 1: x = [C[i][n-1] / C[i][i] for i in range(n)] print("The solution is:", x) else: print("No unique solution exists.") MODIFICATION: Create a new Python file and modify the code to compute the matrix inverse.
PYTHON CODE MODIFICATION The below code is used for solving a system of linear equations using Gaus-Jordan Elimination: # Define the matrix A and vector b A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]] b = [[1], [-1], [2]] # Combine A and b into augmented matrix C C = [row_a + row_b for row_a, row_b in zip(A, b)] # Get the number of rows and columns of the matrix n = len(C) # Set E to 1 (initially assume a unique solution exists) E=1 # Iterate through columns of matrix for j in range(n): # Find the row with the largest magnitude value in column j, only looking at rows j and below pivot_row = max(range(j, n), key=lambda i: abs(C[i][j])) p = C[pivot_row][j] # Check if the pivot is zero if p == 0: E=0 break # If the pivot is not in row j, swap rows j and the pivot row if pivot_row !=j: C[j], C[pivot_row] = C[pivot_row], C[j] # Divide row j by pivot value C[j] = [C[j][i] / p for i in range(n)] # Subtract the pivot row from all other rows to eliminate the value at column j for i in range(n): if i != j: factor = C[i][j] C[i] = [C[i][k] - factor * C[j][k] for k in range(n)] # Check if a unique solution was found if E == 1: x = [C[i][n-1] / C[i][i] for i in range(n)] print("The solution is:", x) else: print("No unique solution exists.") MODIFICATION: Create a new Python file and modify the code to compute the matrix inverse.
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![PYTHON CODE MODIFICATION
The below code is used for solving a system of linear equations using Gaus-Jordan Elimination:
# Define the matrix A and vector b
A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]]
b = [[1], [-1], [2]]
# Combine A and b into augmented matrix C
C = [row_a + row_b for row_a, row_b in zip(A, b)]
# Get the number of rows and columns of the matrix
n = len(C)
# Set E to 1 (initially assume a unique solution exists)
E = 1
# Iterate through columns of matrix
for j in range(n):
# Find the row with the largest magnitude value in column j, only looking at rows j and below
pivot_row = max(range(j, n), key-lambda i: abs(C[i][j]))
p = C[pivot_row][j]
# Check if the pivot is zero
if p == 0:
E = 0
break
# If the pivot is not in row j, swap rows j and the pivot row
if pivot_row != j:
C[j], C[pivot_row] = C[pivot_row], C[j]
# Divide row j by pivot value
C[j] = [C[j][i] / p for i in range(n)]
# Subtract the pivot row from all other rows to eliminate the value at column j
for i in range(n):
if i != j:
factor = C[i][j]
C[i] = [C[i][k] - factor * C[j][k] for k in range(n)]
# Check if a unique solution was found
if E == 1:
x = [C[i][n-1] / C[i][i] for i in range(n)]
print("The solution is:", x)
else:
print("No unique solution exists.")
MODIFICATION:
Create a new Python file and modify the code to compute the matrix inverse.
THANK YOU!!!!! If your solution helps I will upvote!!!!!!!
(PLEASE DO NOT COPY AND PASTE FROM ANOTHER SOLUTION)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0143c836-8f2c-4043-b7b3-58e093f2fd4b%2Fe771051a-3f3d-4c36-97d2-428f16f51d2a%2Fmpz57i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:PYTHON CODE MODIFICATION
The below code is used for solving a system of linear equations using Gaus-Jordan Elimination:
# Define the matrix A and vector b
A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]]
b = [[1], [-1], [2]]
# Combine A and b into augmented matrix C
C = [row_a + row_b for row_a, row_b in zip(A, b)]
# Get the number of rows and columns of the matrix
n = len(C)
# Set E to 1 (initially assume a unique solution exists)
E = 1
# Iterate through columns of matrix
for j in range(n):
# Find the row with the largest magnitude value in column j, only looking at rows j and below
pivot_row = max(range(j, n), key-lambda i: abs(C[i][j]))
p = C[pivot_row][j]
# Check if the pivot is zero
if p == 0:
E = 0
break
# If the pivot is not in row j, swap rows j and the pivot row
if pivot_row != j:
C[j], C[pivot_row] = C[pivot_row], C[j]
# Divide row j by pivot value
C[j] = [C[j][i] / p for i in range(n)]
# Subtract the pivot row from all other rows to eliminate the value at column j
for i in range(n):
if i != j:
factor = C[i][j]
C[i] = [C[i][k] - factor * C[j][k] for k in range(n)]
# Check if a unique solution was found
if E == 1:
x = [C[i][n-1] / C[i][i] for i in range(n)]
print("The solution is:", x)
else:
print("No unique solution exists.")
MODIFICATION:
Create a new Python file and modify the code to compute the matrix inverse.
THANK YOU!!!!! If your solution helps I will upvote!!!!!!!
(PLEASE DO NOT COPY AND PASTE FROM ANOTHER SOLUTION)
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