Mixing
How to reduce matrix into row echelon form in Python
0
$begingroup$
I'm working on a linear algebra homework for a data science class. I'm suppose to make this matrix into row echelon form but I'm stuck.
Here's the current output
I would like to get rid of -0.75, 0.777, and 1.333 in A[2,0], A[3,0], and A[3,1] respectively; they should be zeroed out.
Below is my current code... can anybody please nudge me in the right direction and tell me what step I'm missing?
import numpy as np
def fixRowTwo(A) :
# Sets the sub-diagonal elements of row two to zero
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
# Test if diagonal element is not zero.
if A[2,2] == 0 :
# Add a lower row to row two.
A[2] = A[2] + A[3]
# Sets the sub-diagonal elements to zero again ???
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
if A[2,2] == 0 :
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[2] = A[2] / A[2,2]
return A
def fixRowThree(A) :
# Sets the sub-diagonal elements of row two to zero
A[3] = A[3] - A[3,0] * A[2]
A[3] = A[3] - A[3,1] * A[2]
A[3] = A[3] - A[3,2] * A[2]
# Test if diagonal element is not zero.
if A[3,3] == 0:
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[3] = A[3] / A[3,3]
return A
A = np.array([
[1, 7, 4, 3],
[0, 1, 2, 3],
[3, 2, 0, 3],
[1, 3, 1, 3]
], dtype=np.float_)
fixRowTwo(A)
print("")
print("Row Two:")
print(A)
fixRowThree(A)
print("")
print("Row Three:")
print(A)
linear-algebra python
asked Jan 14 at 10:37
DdflowwDdfloww
32
$endgroup$
add a comment |
0
$begingroup$
I'm working on a linear algebra homework for a data science class. I'm suppose to make this matrix into row echelon form but I'm stuck.
Here's the current output
I would like to get rid of -0.75, 0.777, and 1.333 in A[2,0], A[3,0], and A[3,1] respectively; they should be zeroed out.
Below is my current code... can anybody please nudge me in the right direction and tell me what step I'm missing?
import numpy as np
def fixRowTwo(A) :
# Sets the sub-diagonal elements of row two to zero
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
# Test if diagonal element is not zero.
if A[2,2] == 0 :
# Add a lower row to row two.
A[2] = A[2] + A[3]
# Sets the sub-diagonal elements to zero again ???
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
if A[2,2] == 0 :
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[2] = A[2] / A[2,2]
return A
def fixRowThree(A) :
# Sets the sub-diagonal elements of row two to zero
A[3] = A[3] - A[3,0] * A[2]
A[3] = A[3] - A[3,1] * A[2]
A[3] = A[3] - A[3,2] * A[2]
# Test if diagonal element is not zero.
if A[3,3] == 0:
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[3] = A[3] / A[3,3]
return A
A = np.array([
[1, 7, 4, 3],
[0, 1, 2, 3],
[3, 2, 0, 3],
[1, 3, 1, 3]
], dtype=np.float_)
fixRowTwo(A)
print("")
print("Row Two:")
print(A)
fixRowThree(A)
print("")
print("Row Three:")
print(A)
linear-algebra python
asked Jan 14 at 10:37
DdflowwDdfloww
32
$endgroup$
add a comment |
0
0
0
2
$begingroup$
I'm working on a linear algebra homework for a data science class. I'm suppose to make this matrix into row echelon form but I'm stuck.
Here's the current output
I would like to get rid of -0.75, 0.777, and 1.333 in A[2,0], A[3,0], and A[3,1] respectively; they should be zeroed out.
Below is my current code... can anybody please nudge me in the right direction and tell me what step I'm missing?
import numpy as np
def fixRowTwo(A) :
# Sets the sub-diagonal elements of row two to zero
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
# Test if diagonal element is not zero.
if A[2,2] == 0 :
# Add a lower row to row two.
A[2] = A[2] + A[3]
# Sets the sub-diagonal elements to zero again ???
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
if A[2,2] == 0 :
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[2] = A[2] / A[2,2]
return A
def fixRowThree(A) :
# Sets the sub-diagonal elements of row two to zero
A[3] = A[3] - A[3,0] * A[2]
A[3] = A[3] - A[3,1] * A[2]
A[3] = A[3] - A[3,2] * A[2]
# Test if diagonal element is not zero.
if A[3,3] == 0:
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[3] = A[3] / A[3,3]
return A
A = np.array([
[1, 7, 4, 3],
[0, 1, 2, 3],
[3, 2, 0, 3],
[1, 3, 1, 3]
], dtype=np.float_)
fixRowTwo(A)
print("")
print("Row Two:")
print(A)
fixRowThree(A)
print("")
print("Row Three:")
print(A)
linear-algebra python
asked Jan 14 at 10:37
DdflowwDdfloww
32
$endgroup$
I'm working on a linear algebra homework for a data science class. I'm suppose to make this matrix into row echelon form but I'm stuck.
Here's the current output
I would like to get rid of -0.75, 0.777, and 1.333 in A[2,0], A[3,0], and A[3,1] respectively; they should be zeroed out.
Below is my current code... can anybody please nudge me in the right direction and tell me what step I'm missing?
import numpy as np
def fixRowTwo(A) :
# Sets the sub-diagonal elements of row two to zero
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
# Test if diagonal element is not zero.
if A[2,2] == 0 :
# Add a lower row to row two.
A[2] = A[2] + A[3]
# Sets the sub-diagonal elements to zero again ???
A[2] = A[2] - A[2,0] * A[1]
A[2] = A[2] - A[2,1] * A[1]
if A[2,2] == 0 :
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[2] = A[2] / A[2,2]
return A
def fixRowThree(A) :
# Sets the sub-diagonal elements of row two to zero
A[3] = A[3] - A[3,0] * A[2]
A[3] = A[3] - A[3,1] * A[2]
A[3] = A[3] - A[3,2] * A[2]
# Test if diagonal element is not zero.
if A[3,3] == 0:
print("S I N G U L A R")
sys.Exit()
# Set the diagonal element to one
A[3] = A[3] / A[3,3]
return A
A = np.array([
[1, 7, 4, 3],
[0, 1, 2, 3],
[3, 2, 0, 3],
[1, 3, 1, 3]
], dtype=np.float_)
fixRowTwo(A)
print("")
print("Row Two:")
print(A)
fixRowThree(A)
print("")
print("Row Three:")
print(A)
linear-algebra python
linear-algebra python
asked Jan 14 at 10:37
DdflowwDdfloww
32
asked Jan 14 at 10:37
DdflowwDdfloww
32
asked Jan 14 at 10:37
DdflowwDdfloww
32
asked Jan 14 at 10:37
DdflowwDdfloww
32
asked Jan 14 at 10:37
DdflowwDdfloww
32
32
add a comment |
add a comment |
1 Answer
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$begingroup$
Since the Gaussian process is recursive, we utilize it in the code.
import numpy as np
def row_echelon(A):
""" Return Row Echelon Form of matrix A """
# if matrix A has no columns or rows,
# it is already in REF, so we return itself
r, c = A.shape
if r == 0 or c == 0:
return A
# we search for non-zero element in the first column
for i in range(len(A)):
if A[i,0] != 0:
break
else:
# if all elements in the first column is zero,
# we perform REF on matrix from second column
B = row_echelon(A[:,1:])
# and then add the first zero-column back
return np.hstack([A[:,:1], B])
# if non-zero element happens not in the first row,
# we switch rows
if i > 0:
ith_row = A[i].copy()
A[i] = A[0]
A[0] = ith_row
# we divide first row by first element in it
A[0] = A[0] / A[0,0]
# we subtract all subsequent rows with first row (it has 1 now as first element)
# multiplied by the corresponding element in the first column
A[1:] -= A[0] * A[1:,0:1]
# we perform REF on matrix from second row, from second column
B = row_echelon(A[1:,1:])
# we add first row and first (zero) column, and return
return np.vstack([A[:1], np.hstack([A[1:,:1], B]) ])
A = np.array([[4, 7, 3, 8],
[8, 3, 8, 7],
[2, 9, 5, 3]], dtype='float')
row_echelon(A)
Feel free to add something like print(A[1:,:1]) if you don't understand some of the constructions
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since the Gaussian process is recursive, we utilize it in the code.
import numpy as np
def row_echelon(A):
""" Return Row Echelon Form of matrix A """
# if matrix A has no columns or rows,
# it is already in REF, so we return itself
r, c = A.shape
if r == 0 or c == 0:
return A
# we search for non-zero element in the first column
for i in range(len(A)):
if A[i,0] != 0:
break
else:
# if all elements in the first column is zero,
# we perform REF on matrix from second column
B = row_echelon(A[:,1:])
# and then add the first zero-column back
return np.hstack([A[:,:1], B])
# if non-zero element happens not in the first row,
# we switch rows
if i > 0:
ith_row = A[i].copy()
A[i] = A[0]
A[0] = ith_row
# we divide first row by first element in it
A[0] = A[0] / A[0,0]
# we subtract all subsequent rows with first row (it has 1 now as first element)
# multiplied by the corresponding element in the first column
A[1:] -= A[0] * A[1:,0:1]
# we perform REF on matrix from second row, from second column
B = row_echelon(A[1:,1:])
# we add first row and first (zero) column, and return
return np.vstack([A[:1], np.hstack([A[1:,:1], B]) ])
A = np.array([[4, 7, 3, 8],
[8, 3, 8, 7],
[2, 9, 5, 3]], dtype='float')
row_echelon(A)
Feel free to add something like print(A[1:,:1]) if you don't understand some of the constructions
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
$endgroup$
add a comment |
0
$begingroup$
Since the Gaussian process is recursive, we utilize it in the code.
import numpy as np
def row_echelon(A):
""" Return Row Echelon Form of matrix A """
# if matrix A has no columns or rows,
# it is already in REF, so we return itself
r, c = A.shape
if r == 0 or c == 0:
return A
# we search for non-zero element in the first column
for i in range(len(A)):
if A[i,0] != 0:
break
else:
# if all elements in the first column is zero,
# we perform REF on matrix from second column
B = row_echelon(A[:,1:])
# and then add the first zero-column back
return np.hstack([A[:,:1], B])
# if non-zero element happens not in the first row,
# we switch rows
if i > 0:
ith_row = A[i].copy()
A[i] = A[0]
A[0] = ith_row
# we divide first row by first element in it
A[0] = A[0] / A[0,0]
# we subtract all subsequent rows with first row (it has 1 now as first element)
# multiplied by the corresponding element in the first column
A[1:] -= A[0] * A[1:,0:1]
# we perform REF on matrix from second row, from second column
B = row_echelon(A[1:,1:])
# we add first row and first (zero) column, and return
return np.vstack([A[:1], np.hstack([A[1:,:1], B]) ])
A = np.array([[4, 7, 3, 8],
[8, 3, 8, 7],
[2, 9, 5, 3]], dtype='float')
row_echelon(A)
Feel free to add something like print(A[1:,:1]) if you don't understand some of the constructions
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
$endgroup$
add a comment |
0
0
0
$begingroup$
Since the Gaussian process is recursive, we utilize it in the code.
import numpy as np
def row_echelon(A):
""" Return Row Echelon Form of matrix A """
# if matrix A has no columns or rows,
# it is already in REF, so we return itself
r, c = A.shape
if r == 0 or c == 0:
return A
# we search for non-zero element in the first column
for i in range(len(A)):
if A[i,0] != 0:
break
else:
# if all elements in the first column is zero,
# we perform REF on matrix from second column
B = row_echelon(A[:,1:])
# and then add the first zero-column back
return np.hstack([A[:,:1], B])
# if non-zero element happens not in the first row,
# we switch rows
if i > 0:
ith_row = A[i].copy()
A[i] = A[0]
A[0] = ith_row
# we divide first row by first element in it
A[0] = A[0] / A[0,0]
# we subtract all subsequent rows with first row (it has 1 now as first element)
# multiplied by the corresponding element in the first column
A[1:] -= A[0] * A[1:,0:1]
# we perform REF on matrix from second row, from second column
B = row_echelon(A[1:,1:])
# we add first row and first (zero) column, and return
return np.vstack([A[:1], np.hstack([A[1:,:1], B]) ])
A = np.array([[4, 7, 3, 8],
[8, 3, 8, 7],
[2, 9, 5, 3]], dtype='float')
row_echelon(A)
Feel free to add something like print(A[1:,:1]) if you don't understand some of the constructions
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
$endgroup$
Since the Gaussian process is recursive, we utilize it in the code.
import numpy as np
def row_echelon(A):
""" Return Row Echelon Form of matrix A """
# if matrix A has no columns or rows,
# it is already in REF, so we return itself
r, c = A.shape
if r == 0 or c == 0:
return A
# we search for non-zero element in the first column
for i in range(len(A)):
if A[i,0] != 0:
break
else:
# if all elements in the first column is zero,
# we perform REF on matrix from second column
B = row_echelon(A[:,1:])
# and then add the first zero-column back
return np.hstack([A[:,:1], B])
# if non-zero element happens not in the first row,
# we switch rows
if i > 0:
ith_row = A[i].copy()
A[i] = A[0]
A[0] = ith_row
# we divide first row by first element in it
A[0] = A[0] / A[0,0]
# we subtract all subsequent rows with first row (it has 1 now as first element)
# multiplied by the corresponding element in the first column
A[1:] -= A[0] * A[1:,0:1]
# we perform REF on matrix from second row, from second column
B = row_echelon(A[1:,1:])
# we add first row and first (zero) column, and return
return np.vstack([A[:1], np.hstack([A[1:,:1], B]) ])
A = np.array([[4, 7, 3, 8],
[8, 3, 8, 7],
[2, 9, 5, 3]], dtype='float')
row_echelon(A)
Feel free to add something like print(A[1:,:1]) if you don't understand some of the constructions
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
answered Jan 14 at 11:19
Vasily MitchVasily Mitch
2,3241311
2,3241311
add a comment |
add a comment |
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