PPPK_programs_REF_and

A = matrix (RR, [[4, -1, -1, 0, -1], 
                 [-1, 4, -1, -1, 0], 
                 [-1, -1, 4, -1, -1], 
                 [0, -1, -1, 4, -1], 
                 [-1, 0, -1, -1, 4]])
show (A)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \end{array}\right)$

b = vector (RR, [5, -5, 4, -6, -5])
show (b.column())

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{r} 5.00000000000000 \\ -5.00000000000000 \\ 4.00000000000000 \\ -6.00000000000000 \\ -5.00000000000000 \end{array}\right)$

Ab = block_matrix ([[A, b.column()]])
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

A.rescale_row?

File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.rescale_row(i, s, start_col=0)

Docstring:

Replace i-th row of self by s times i-th row of A.

INPUT:

i - ith row

s - scalar

start_col - only rescale entries at this column and to the right

EXAMPLES: We rescale the second row of a matrix over the rational numbers:
sage: a = matrix(QQ,3,range(6)); a
[0 1]
[2 3]
[4 5]
sage: a.rescale_row(1,1/2); a
[ 0   1]
[ 1 3/2]
[ 4   5]
We rescale the second row of a matrix over a polynomial ring:
sage: R.<x> = QQ[]
sage: a = matrix(R,3,[1,x,x^2,x^3,x^4,x^5]);a
[  1   x]
[x^2 x^3]
[x^4 x^5]
sage: a.rescale_row(1,1/2); a
[      1       x]
[1/2*x^2 1/2*x^3]
[    x^4     x^5]
We try and fail to rescale a matrix over the integers by a non-integer:
sage: a = matrix(ZZ,2,3,[0,1,2, 3,4,4]); a
[0 1 2]
[3 4 4]
sage: a.rescale_row(1,1/2)
Traceback (click to the left of this block for traceback)
...

                                
                            
File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.rescale_row(i, s, start_col=0)

Docstring:
Replace i-th row of self by s times i-th row of A.

INPUT:

i - ith row

s - scalar

start_col - only rescale entries at this column and to the right

EXAMPLES: We rescale the second row of a matrix over the rational numbers:
sage: a = matrix(QQ,3,range(6)); a
[0 1]
[2 3]
[4 5]
sage: a.rescale_row(1,1/2); a
[ 0   1]
[ 1 3/2]
[ 4   5]
We rescale the second row of a matrix over a polynomial ring:
sage: R.<x> = QQ[]
sage: a = matrix(R,3,[1,x,x^2,x^3,x^4,x^5]);a
[  1   x]
[x^2 x^3]
[x^4 x^5]
sage: a.rescale_row(1,1/2); a
[      1       x]
[1/2*x^2 1/2*x^3]
[    x^4     x^5]
We try and fail to rescale a matrix over the integers by a non-integer:
sage: a = matrix(ZZ,2,3,[0,1,2, 3,4,4]); a
[0 1 2]
[3 4 4]
sage: a.rescale_row(1,1/2)
Traceback (most recent call last):
...
TypeError: Rescaling row by Rational Field element cannot be done over Integer Ring, use change_ring or with_rescaled_row instead.
To rescale the matrix by 1/2, you must change the base ring to the rationals:
sage: a = a.change_ring(QQ); a
[0 1 2]
[3 4 4]
sage: a.rescale_col(1,1/2); a
[  0 1/2   2]
[  3   2   4]

Ab.rescale_row (0, 1/Ab[0, 0])
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

A.add_multiple_of_row?

File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.add_multiple_of_row(i, j, s, start_col=0)

Docstring:

Add s times row j to row i.

EXAMPLES: We add -3 times the first row to the second row of an integer matrix, remembering to start numbering rows at zero:
sage: a = matrix(ZZ,2,3,range(6)); a
[0 1 2]
[3 4 5]
sage: a.add_multiple_of_row(1,0,-3)
sage: a
[ 0  1  2]
[ 3  1 -1]
To add a rational multiple, we first need to change the base ring:
sage: a = a.change_ring(QQ)
sage: a.add_multiple_of_row(1,0,1/3)
sage: a
[   0    1    2]
[   3  4/3 -1/3]
If not, we get an error message:
sage: a.add_multiple_of_row(1,0,i)
Traceback (click to the left of this block for traceback)
...

                                
                            
File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.add_multiple_of_row(i, j, s, start_col=0)

Docstring:
Add s times row j to row i.

EXAMPLES: We add -3 times the first row to the second row of an integer matrix, remembering to start numbering rows at zero:
sage: a = matrix(ZZ,2,3,range(6)); a
[0 1 2]
[3 4 5]
sage: a.add_multiple_of_row(1,0,-3)
sage: a
[ 0  1  2]
[ 3  1 -1]
To add a rational multiple, we first need to change the base ring:
sage: a = a.change_ring(QQ)
sage: a.add_multiple_of_row(1,0,1/3)
sage: a
[   0    1    2]
[   3  4/3 -1/3]
If not, we get an error message:
sage: a.add_multiple_of_row(1,0,i)
Traceback (most recent call last):
...
TypeError: Multiplying row by Symbolic Ring element cannot be done over Rational Field, use change_ring or with_added_multiple_of_row instead.

for i in xrange (1, Ab.nrows()) :
    Ab.add_multiple_of_row (i, 0, -Ab[i, 0])
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 3.75000000000000 & -1.25000000000000 & -1.00000000000000 & -0.250000000000000 & -3.75000000000000 \\ 0.000000000000000 & -1.25000000000000 & 3.75000000000000 & -1.00000000000000 & -1.25000000000000 & 5.25000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ 0.000000000000000 & -0.250000000000000 & -1.25000000000000 & -1.00000000000000 & 3.75000000000000 & -3.75000000000000 \end{array}\right)$

Ab.rescale_row (1, 1/Ab[1,1], 1)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & -1.25000000000000 & 3.75000000000000 & -1.00000000000000 & -1.25000000000000 & 5.25000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ 0.000000000000000 & -0.250000000000000 & -1.25000000000000 & -1.00000000000000 & 3.75000000000000 & -3.75000000000000 \end{array}\right)$

for i in xrange (2, Ab.nrows()) :
    Ab.add_multiple_of_row (i, 1, -Ab[i, 1], 1)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 3.33333333333333 & -1.33333333333333 & -1.33333333333333 & 4.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & 3.73333333333333 & -1.06666666666667 & -7.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & -1.06666666666667 & 3.73333333333333 & -4.00000000000000 \end{array}\right)$

Ab.rescale_row (2, 1/Ab[2,2], 2)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & 3.73333333333333 & -1.06666666666667 & -7.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & -1.06666666666667 & 3.73333333333333 & -4.00000000000000 \end{array}\right)$

for i in xrange (3, Ab.nrows()) :
    Ab.add_multiple_of_row (i, 2, -Ab[i, 2], 2)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 3.20000000000000 & -1.60000000000000 & -5.40000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab.rescale_row (3, 1/Ab[3, 3], 3)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & -1.68750000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

for i in xrange (4, Ab.nrows()) :
    Ab.add_multiple_of_row (i, 3, -Ab[i, 3], 3)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & -1.68750000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.40000000000000 & -5.10000000000000 \end{array}\right)$

Ab.rescale_row (4, 1/Ab[4, 4], 4)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & -1.68750000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.12500000000000 \end{array}\right)$

Ab = block_matrix ([[A, b.column()]])
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

for j in xrange (0, Ab.nrows()) :
    Ab.rescale_row (j, 1/Ab[j, j], j)
    for i in xrange (j + 1, Ab.nrows()) :
        Ab.add_multiple_of_row (i, j, -Ab[i, j], j)
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & -1.68750000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.12500000000000 \end{array}\right)$

xx = vector (RR, 5)
xx[4] = Ab[4,5]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0.000000000000000,\,0.000000000000000,\,0.000000000000000,\,0.000000000000000,\,-2.12500000000000\right)$

xx[3] = Ab[3,5] - Ab[3,4] * xx[4]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0.000000000000000,\,0.000000000000000,\,0.000000000000000,\,-2.75000000000000,\,-2.12500000000000\right)$

xx[2] = Ab[2,5] - Ab[2,4] * xx[4] - Ab[2,3] * xx[3]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0.000000000000000,\,0.000000000000000,\,-0.749999999999999,\,-2.75000000000000,\,-2.12500000000000\right)$

xx[1] = Ab[1,5]
for i in xrange (Ab.ncols() - 2, 1, -1) :
    xx[1] -= Ab[1,i] * xx[i]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0.000000000000000,\,-2.12500000000000,\,-0.749999999999999,\,-2.75000000000000,\,-2.12500000000000\right)$

xx[0] = Ab[0,5]
for i in xrange (Ab.ncols() - 2, 0, -1) :
    xx[0] -= Ab[0,i] * xx[i]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(3.33066907387547 \times 10^{-16},\,-2.12500000000000,\,-0.749999999999999,\,-2.75000000000000,\,-2.12500000000000\right)$

xx = vector (RR, 5)
for j in xrange (len (xx) - 1, -1, -1) :
    xx[j] = Ab[j, Ab.ncols() - 1]
    for i in xrange (Ab.ncols() - 2, j, -1) :
        xx[j] -= Ab[j, i] * xx[i]
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(3.33066907387547 \times 10^{-16},\,-2.12500000000000,\,-0.749999999999999,\,-2.75000000000000,\,-2.12500000000000\right)$

def chop (tolerance = None) :
    def f_ (x) :
        tol = sqrt ((parent(x)(1)).ulp()) if tolerance == None else tolerance
        return x if (abs(x) >= tol) else parent(x)(0)
    return f_

show (map (chop(), xx))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.000000000000000, -2.12500000000000, -0.749999999999999, -2.75000000000000, -2.12500000000000\right]$

show (A * xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-6.00000000000000,\,-5.00000000000000\right)$

show (b)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-6.00000000000000,\,-5.00000000000000\right)$

Ab2 = block_matrix ([[A, b.column()]])
Ab2[3] = -Ab2[0]/3 - Ab2[2]/7
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ -1.19047619047619 & 0.476190476190476 & -0.238095238095238 & 0.142857142857143 & 0.476190476190476 & -2.23809523809524 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

A2 = Ab2[:, 0:5]
show (A2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 \\ -1.19047619047619 & 0.476190476190476 & -0.238095238095238 & 0.142857142857143 & 0.476190476190476 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \end{array}\right)$

b2 = Ab2.column (5)
show (b2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

for j in xrange (0, Ab2.nrows()) :
    Ab2.rescale_row (j, 1/Ab2[j, j], j)
    for i in xrange (j + 1, Ab2.nrows()) :
        Ab2.add_multiple_of_row (i, j, -Ab2[i, j], j)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.750000000000000 \end{array}\right)$

xx2 = vector (RR, 5)
for j in xrange (len (xx2) - 1, -1, -1) :
    xx2[j] = Ab2[j, Ab2.ncols() - 1]
    for i in xrange (Ab2.ncols() - 2, j, -1) :
        xx2[j] -= Ab2[j, i] * xx2[i]
show (xx2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.10000000000000,\,-0.750000000000000,\,0.900000000000000,\,0.000000000000000,\,-0.750000000000000\right)$

show (A2 * xx2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

show (b2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

Ab2 = block_matrix ([[A2, b2.column()]])
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ -1.19047619047619 & 0.476190476190476 & -0.238095238095238 & 0.142857142857143 & 0.476190476190476 & -2.23809523809524 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
Ab2.rescale_row (0, 1/Ab2[0, 0])
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ -1.19047619047619 & 0.476190476190476 & -0.238095238095238 & 0.142857142857143 & 0.476190476190476 & -2.23809523809524 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
for i in xrange (1, Ab2.nrows()) :
    Ab2.add_multiple_of_row (i, 0, -Ab2[i, 0])
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 3.75000000000000 & -1.25000000000000 & -1.00000000000000 & -0.250000000000000 & -3.75000000000000 \\ 0.000000000000000 & -1.25000000000000 & 3.75000000000000 & -1.00000000000000 & -1.25000000000000 & 5.25000000000000 \\ 0.000000000000000 & 0.178571428571429 & -0.535714285714286 & 0.142857142857143 & 0.178571428571429 & -0.750000000000000 \\ 0.000000000000000 & -0.250000000000000 & -1.25000000000000 & -1.00000000000000 & 3.75000000000000 & -3.75000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
Ab2.rescale_row (1, 1/Ab2[1, 1], 1)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & -1.25000000000000 & 3.75000000000000 & -1.00000000000000 & -1.25000000000000 & 5.25000000000000 \\ 0.000000000000000 & 0.178571428571429 & -0.535714285714286 & 0.142857142857143 & 0.178571428571429 & -0.750000000000000 \\ 0.000000000000000 & -0.250000000000000 & -1.25000000000000 & -1.00000000000000 & 3.75000000000000 & -3.75000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
for i in xrange (2, Ab2.nrows()) :
    Ab2.add_multiple_of_row (i, 1, -Ab2[i, 1], 1)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 3.33333333333333 & -1.33333333333333 & -1.33333333333333 & 4.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -0.476190476190476 & 0.190476190476190 & 0.190476190476190 & -0.571428571428571 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & -1.06666666666667 & 3.73333333333333 & -4.00000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
Ab2.rescale_row (2, 1/Ab2[2, 2], 2)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & -0.476190476190476 & 0.190476190476190 & 0.190476190476190 & -0.571428571428571 \\ 0.000000000000000 & 0.000000000000000 & -1.33333333333333 & -1.06666666666667 & 3.73333333333333 & -4.00000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
for i in xrange (3, Ab2.nrows()) :
    Ab2.add_multiple_of_row (i, 2, -Ab2[i, 2], 2)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.77555756156289 \times 10^{-17} & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2c = copy (Ab2)
Ab2.rescale_row (3, 1/Ab2[3, 3], 3)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

for i in xrange (4, Ab2.nrows()) :
    Ab2.add_multiple_of_row (i, 3, -Ab2[i, 3], 3)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2.rescale_row (4, 1/Ab2[4, 4], 4)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.750000000000000 \end{array}\right)$

Ab2 = copy (Ab2c)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.77555756156289 \times 10^{-17} & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2[3, 3] = 0.
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2.rescale_row (3, 1/Ab2[3, 3], 3)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & NaN & NaN & NaN \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2[3, 3] = Ab2[3, 4] = Ab2[3, 5] = 0.
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

A.swap_rows?

File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.swap_rows(r1, r2)

Docstring:

Swap rows r1 and r2 of A.

EXAMPLES: We create a rational matrix:
sage: M = MatrixSpace(QQ,3,3)
sage: A = M([1,9,-7,4/5,4,3,6,4,3])
sage: A
[  1   9  -7]
[4/5   4   3]
[  6   4   3]
Since the first row is numbered zero, this swaps the first and third rows:
sage: A.swap_rows(0,2); A
[  6   4   3]
[4/5   4   3]
[  1   9  -7]

File: /opt/SageMath/local/lib/python2.7/site-packages/sage/matrix/matrix0.pyx

Type: <type ‘builtin_function_or_method’>

Definition: A.swap_rows(r1, r2)

Docstring:

Swap rows r1 and r2 of A.

EXAMPLES: We create a rational matrix:
sage: M = MatrixSpace(QQ,3,3)
sage: A = M([1,9,-7,4/5,4,3,6,4,3])
sage: A
[  1   9  -7]
[4/5   4   3]
[  6   4   3]
Since the first row is numbered zero, this swaps the first and third rows:
sage: A.swap_rows(0,2); A
[  6   4   3]
[4/5   4   3]
[  1   9  -7]

Ab2.swap_rows (3, 4)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

Ab2.rescale_row (3, 1/Ab2[3, 3], 3)
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

xx2cc = vector (RR, 5)
for j in xrange (len (xx2cc) - 1, -1, -1) :
    xx2cc[j] = Ab2[j, Ab2.ncols() - 1]
    for i in xrange (Ab2.ncols() - 2, j, -1) :
        xx2cc[j] -= Ab2[j, i] * xx2cc[i]
show (xx2cc)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,0.000000000000000\right)$

show (Ab2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.77555756156289 \times 10^{-17} & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \end{array}\right)$

Ab2c.swap_rows (3, 4)
show (Ab2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & -1.60000000000000 & 3.20000000000000 & -2.40000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.77555756156289 \times 10^{-17} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

Ab2c.rescale_row (3, 1/Ab2c[3, 3], 3)
show (Ab2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 2.77555756156289 \times 10^{-17} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

for i in xrange (4, Ab2c.nrows()) :
    Ab2c.add_multiple_of_row (i, 3, -Ab2c[i, 3], 3)
show (Ab2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 3.08148791101958 \times 10^{-33} & 5.55111512312578 \times 10^{-17} & -4.16333634234434 \times 10^{-17} \end{array}\right)$

show (Ab2c.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

xx2c = vector (RR, 5)
for j in xrange (len (xx2c) - 1, -1, -1) :
    xx2c[j] = Ab2c[j, Ab2c.ncols() - 1]
    for i in xrange (Ab2c.ncols() - 2, j, -1) :
        xx2c[j] -= Ab2c[j, i] * xx2c[i]
show (xx2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,-4.16333634234434 \times 10^{-17}\right)$

show (xx2cc)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,0.000000000000000\right)$

show (map (chop(), xx2c))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.70000000000000, 0.000000000000000, 1.80000000000000, 1.50000000000000, 0.000000000000000\right]$

show (map (chop(), xx2cc))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.70000000000000, 0.000000000000000, 1.80000000000000, 1.50000000000000, 0.000000000000000\right]$

show (A2 * xx2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

show (A2 * xx2cc)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

show (b2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-2.23809523809524,\,-5.00000000000000\right)$

show (xx2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.10000000000000,\,-0.750000000000000,\,0.900000000000000,\,0.000000000000000,\,-0.750000000000000\right)$

xxs = [var ('xx%d' % i) for i in xrange (5)]
yys = solve (map (lambda x, y : x == y, A2 * vector (xxs), b2), xxs)
show (yys)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[\mathit{xx}_{0} = \frac{4}{5} \, r_{2} + \frac{17}{10}, \mathit{xx}_{1} = r_{2}, \mathit{xx}_{2} = \frac{6}{5} \, r_{2} + \frac{9}{5}, \mathit{xx}_{3} = 2 \, r_{2} + \frac{3}{2}, \mathit{xx}_{4} = r_{2}\right]\right]$

show (map (lambda x : x.rhs().subs (r2 = -0.75), yys[0]))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.10000000000000, -0.750000000000000, 0.900000000000000, 0.000000000000000, -0.750000000000000\right]$

def is_zero (x) :
    return x == 0

def is_zero_wtol (tolerance = None) :
    def f_ (x) :
        t = sqrt ((parent(x)(1)).ulp()) if tolerance == None else tolerance
        return abs (x) <= t
    return f_

is_zero (0.)

True

True

is_zero_wtol() (0.)

True

True

u53 = (1.).ulp()
u53

2.22044604925031e-16

2.22044604925031e-16

is_zero (u53)

False

False

is_zero_wtol() (u53)

True

True

is_zero_wtol() (sqrt (u53))

True

True

is_zero_wtol() (sqrt (u53) + u53)

False

False

is_zero_wtol() (sqrt (u53) + u53^2)

True

True

is_zero_wtol() (-sqrt (u53 + u53^2))

True

True

def ref_wpp (A, is_zero_f = is_zero_wtol(), in_place = False) :
    
    Ar = A if in_place else copy (A)
        
    r = Ar.nrows()
    c = Ar.ncols() 
    pivot = 0
    pivotlist = []
    
    for j in xrange (c) :
        if pivot == r :
            break
        maxrow = pivot
        for k in xrange (pivot + 1, r):
            if abs (Ar[k, j]) > abs (Ar[maxrow, j]) : 
                maxrow = k
        if pivot != maxrow :
            Ar.swap_rows (pivot, maxrow)
        if not is_zero_f (Ar[pivot, j]) :
            scale = Ar[pivot, j]
            Ar.rescale_row (pivot, 1/scale, j)
            for i in xrange (pivot + 1, r) :
                scale = Ar[i, j]
                Ar.add_multiple_of_row (i, pivot, -scale, j)
            pivotlist.append (j)
            pivot += 1
            
    return Ar, pivotlist


def ref_solution (Ar, pivl) :
     
    b = Ar.column (-1)
    nx = Ar.ncols() - 1
    np = len (pivl)
    x = zero_vector (Ar.base_ring(), nx)
    
    for i in xrange (np - 1, -1, -1) :
        ii = pivl[i]
        x[ii] = b[i]
        for j in xrange (np - 1, i, -1) :
            jj = pivl[j]
            x[ii] -= Ar[i, jj] * x[jj]
            
    return x

Ab = block_matrix ([[A, b.column()]])
show (Ab)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -6.00000000000000 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

Abr, pl = ref_wpp (Ab)
show (Abr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & -1.68750000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.12500000000000 \end{array}\right)$

pl

[0, 1, 2, 3, 4]

[0, 1, 2, 3, 4]

xx = ref_solution (Abr, pl)
show (xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(3.33066907387547 \times 10^{-16},\,-2.12500000000000,\,-0.749999999999999,\,-2.75000000000000,\,-2.12500000000000\right)$

show (A * xx)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(5.00000000000000,\,-5.00000000000000,\,4.00000000000000,\,-6.00000000000000,\,-5.00000000000000\right)$

Ab2 = block_matrix ([[A2, b2.column()]])
show (Ab2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & 5.00000000000000 \\ -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 0.000000000000000 & -5.00000000000000 \\ -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 \\ -1.19047619047619 & 0.476190476190476 & -0.238095238095238 & 0.142857142857143 & 0.476190476190476 & -2.23809523809524 \\ -1.00000000000000 & 0.000000000000000 & -1.00000000000000 & -1.00000000000000 & 4.00000000000000 & -5.00000000000000 \end{array}\right)$

Ab2r, pl2 = ref_wpp (Ab2)
show (Ab2r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 3.08148791101958 \times 10^{-33} & 5.55111512312578 \times 10^{-17} & -4.16333634234434 \times 10^{-17} \end{array}\right)$

pl2

[0, 1, 2, 3]

[0, 1, 2, 3]

xx2 = ref_solution (Ab2r, pl2)
show (xx2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,0.000000000000000\right)$

show (xx2c)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,-4.16333634234434 \times 10^{-17}\right)$

show (xx2cc)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.70000000000000,\,1.11022302462516 \times 10^{-16},\,1.80000000000000,\,1.50000000000000,\,0.000000000000000\right)$

Ab2r0, pl20 = ref_wpp (Ab2, is_zero)
show (Ab2r0)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 3.08148791101958 \times 10^{-33} & 1.00000000000000 & -0.750000000000000 \end{array}\right)$

show (Ab2r0.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & -0.250000000000000 & -0.250000000000000 & 0.000000000000000 & -0.250000000000000 & 1.25000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.333333333333333 & -0.266666666666667 & -0.0666666666666667 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.400000000000000 & -0.400000000000000 & 1.20000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -2.00000000000000 & 1.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -0.750000000000000 \end{array}\right)$

show (ref_solution (Ab2r0, pl20))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.10000000000000,\,-0.750000000000000,\,0.900000000000000,\,0.000000000000000,\,-0.750000000000000\right)$

show (map (lambda x : x.rhs().subs (r2 = -0.75), yys[0]))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.10000000000000, -0.750000000000000, 0.900000000000000, 0.000000000000000, -0.750000000000000\right]$

A3 = matrix (RR, [[1, 2, 3],
                  [1, 0, 1], 
                  [2, 4, 0]])
show (A3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1.00000000000000 & 2.00000000000000 & 3.00000000000000 \\ 1.00000000000000 & 0.000000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 0.000000000000000 \end{array}\right)$

b3 = matrix (RR, 3, 1, [1, -3, 8])
show (b3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{r} 1.00000000000000 \\ -3.00000000000000 \\ 8.00000000000000 \end{array}\right)$

Ab3 = block_matrix ([[A3, b3]])
show (Ab3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr|r} 1.00000000000000 & 2.00000000000000 & 3.00000000000000 & 1.00000000000000 \\ 1.00000000000000 & 0.000000000000000 & 1.00000000000000 & -3.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 0.000000000000000 & 8.00000000000000 \end{array}\right)$

Ab3r, pl3 = ref_wpp (Ab3)
show (Ab3r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr|r} 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 4.00000000000000 \\ 0.000000000000000 & 1.00000000000000 & -0.500000000000000 & 3.50000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \end{array}\right)$

pl3

[0, 1, 2]

[0, 1, 2]

xx3 = ref_solution (Ab3r, pl3)
show (xx3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-2.00000000000000,\,3.00000000000000,\,-1.00000000000000\right)$

A4 = matrix (RR, [[1, 2, 2, 3],
                  [2, 4, 1, 3],
                  [3, 6, 1, 4]])
show (A4)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 3.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 1.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 6.00000000000000 & 1.00000000000000 & 4.00000000000000 \end{array}\right)$

b4 = vector (RR, [4, 5, 7])
show (b4)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(4.00000000000000,\,5.00000000000000,\,7.00000000000000\right)$

Ab4 = block_matrix ([[A4, b4.column()]])
show (Ab4)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 3.00000000000000 & 4.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 1.00000000000000 & 3.00000000000000 & 5.00000000000000 \\ 3.00000000000000 & 6.00000000000000 & 1.00000000000000 & 4.00000000000000 & 7.00000000000000 \end{array}\right)$

Ab4r, pl4 = ref_wpp (Ab4)
show (Ab4r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 0.333333333333333 & 1.33333333333333 & 2.33333333333333 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.11022302462516 \times 10^{-16} & 4.99600361081320 \times 10^{-16} \end{array}\right)$

show (Ab4r.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 0.333333333333333 & 1.33333333333333 & 2.33333333333333 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

pl4

[0, 2]

[0, 2]

xx4 = ref_solution (Ab4r, pl4)
show (xx4)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2.00000000000000,\,0.000000000000000,\,1.00000000000000,\,0.000000000000000\right)$

A5 = matrix (RR, [[1, 1, 2, 2, 1],
                  [2, 2, 4, 4, 3],
                  [2, 2, 4, 4, 2],
                  [3, 5, 8, 6, 5],
                  [3, 3, 6, 6, 4]])
show (A5)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1.00000000000000 & 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 3.00000000000000 \\ 2.00000000000000 & 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 2.00000000000000 \\ 3.00000000000000 & 5.00000000000000 & 8.00000000000000 & 6.00000000000000 & 5.00000000000000 \\ 3.00000000000000 & 3.00000000000000 & 6.00000000000000 & 6.00000000000000 & 4.00000000000000 \end{array}\right)$

b5 = vector (RR,[1, 1, 2, 3, 2])
show (b5)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,1.00000000000000,\,2.00000000000000,\,3.00000000000000,\,2.00000000000000\right)$

Ab5 = block_matrix ([[A5, b5.column()]])
show (Ab5)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 3.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 2.00000000000000 & 2.00000000000000 \\ 3.00000000000000 & 5.00000000000000 & 8.00000000000000 & 6.00000000000000 & 5.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 3.00000000000000 & 6.00000000000000 & 6.00000000000000 & 4.00000000000000 & 2.00000000000000 \end{array}\right)$

Ab5r, pl5 = ref_wpp (Ab5)
show (Ab5r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & 1.66666666666667 & 2.66666666666667 & 2.00000000000000 & 1.66666666666667 & 1.00000000000000 \\ 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & -0.000000000000000 & 0.500000000000000 & 0.500000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

pl5

[0, 1, 4]

[0, 1, 4]

xx5 = ref_solution (Ab5r, pl5)
show (xx5)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,1.00000000000000,\,0.000000000000000,\,0.000000000000000,\,-1.00000000000000\right)$

A6 = matrix (RR, [[1, 2, 2, 1],
                  [2, 4, 4, 3],
                  [2, 4, 4, 2],
                  [5, 8, 6, 5],
                  [3, 6, 6, 4]])
show (A6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 3.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 2.00000000000000 \\ 5.00000000000000 & 8.00000000000000 & 6.00000000000000 & 5.00000000000000 \\ 3.00000000000000 & 6.00000000000000 & 6.00000000000000 & 4.00000000000000 \end{array}\right)$

Ab6 = block_matrix ([[A6, b5.column()]])
show (Ab6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 3.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 2.00000000000000 & 2.00000000000000 \\ 5.00000000000000 & 8.00000000000000 & 6.00000000000000 & 5.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 6.00000000000000 & 6.00000000000000 & 4.00000000000000 & 2.00000000000000 \end{array}\right)$

Ab6r, pl6 = ref_wpp (Ab6)
show (Ab6r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 1.60000000000000 & 1.20000000000000 & 1.00000000000000 & 0.600000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.833333333333334 & 0.166666666666667 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 0.000000000000000 & -3.33066907387547 \times 10^{-16} \\ 0.000000000000000 & 0.000000000000000 & -3.33066907387547 \times 10^{-16} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

show (Ab6r.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 1.60000000000000 & 1.20000000000000 & 1.00000000000000 & 0.600000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.833333333333334 & 0.166666666666667 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

pl6

[0, 1, 3]

[0, 1, 3]

xx6 = ref_solution (Ab6r, pl6)
show (xx6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-4.44089209850063 \times 10^{-16},\,1.00000000000000,\,0.000000000000000,\,-1.00000000000000\right)$

show (A6 * xx6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,1.00000000000000,\,2.00000000000000,\,3.00000000000000,\,2.00000000000000\right)$

b6 = vector (RR,[1, 1, 4, 3, 2])
show (b6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,1.00000000000000,\,4.00000000000000,\,3.00000000000000,\,2.00000000000000\right)$

Ab7 = block_matrix ([[A6, b6.column()]])
show (Ab7)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 3.00000000000000 & 1.00000000000000 \\ 2.00000000000000 & 4.00000000000000 & 4.00000000000000 & 2.00000000000000 & 4.00000000000000 \\ 5.00000000000000 & 8.00000000000000 & 6.00000000000000 & 5.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 6.00000000000000 & 6.00000000000000 & 4.00000000000000 & 2.00000000000000 \end{array}\right)$

Ab7r, pl7 = ref_wpp (Ab7)
show (Ab7r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 1.60000000000000 & 1.20000000000000 & 1.00000000000000 & 0.600000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.833333333333334 & 0.166666666666667 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 1.00000000000000 & -4.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -3.33066907387547 \times 10^{-16} & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

show (Ab7r.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 1.60000000000000 & 1.20000000000000 & 1.00000000000000 & 0.600000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.833333333333334 & 0.166666666666667 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -4.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

pl7

[0, 1, 3, 4]

[0, 1, 3, 4]

def rref_wpp (A, is_zero_f = is_zero_wtol(), in_place = False) :
    
    Arr = A if in_place == True else copy (A)
        
    r = Arr.nrows()
    c = Arr.ncols() 
    pivot = 0
    pivotlist = []
    
    for j in xrange (c) :
        if pivot == r :
            break
        maxrow = pivot
        for k in xrange (pivot + 1, r):
            if abs (Arr[k, j]) > abs (Arr[maxrow, j]) : 
                maxrow = k
        if pivot != maxrow :
            Arr.swap_rows (pivot, maxrow)
        if not is_zero_f (Arr[pivot, j]) :
            scale = Arr[pivot, j]
            Arr.rescale_row (pivot, 1/scale, j)
            for i in xrange (r) :
                if i == pivot :
                    continue
                scale = Arr[i, j]
                Arr.add_multiple_of_row (i, pivot, -scale, j)
            pivotlist.append (j)
            pivot += 1
            
    return Arr, pivotlist


def rref_solution (Arr, pivl) :

    b = Arr.column (-1)
    nx = Arr.ncols() - 1
    np = len (pivl)
    x = zero_vector (Arr.base_ring(), nx) 

    for i in xrange (np - 1, -1, -1) :
        x[pivl[i]] = b[i] 

    return x

Ab3rr, ppl3 = rref_wpp (Ab3)
show (Ab3rr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr|r} 1.00000000000000 & 0.000000000000000 & 0.000000000000000 & -2.00000000000000 \\ 0.000000000000000 & 1.00000000000000 & 0.000000000000000 & 3.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \end{array}\right)$

ppl3

[0, 1, 2]

[0, 1, 2]

xx3r = rref_solution (Ab3rr, ppl3)
show (xx3r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-2.00000000000000,\,3.00000000000000,\,-1.00000000000000\right)$

Ab4rr, ppl4 = rref_wpp (Ab4)
show (Ab4rr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 1.00000000000000 & 2.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.11022302462516 \times 10^{-16} & 4.99600361081320 \times 10^{-16} \end{array}\right)$

show (Ab4rr.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 1.00000000000000 & 2.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

ppl4

[0, 2]

[0, 2]

xx4r = rref_solution (Ab4rr, ppl4)
show (xx4r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2.00000000000000,\,0.000000000000000,\,1.00000000000000,\,0.000000000000000\right)$

Ab5rr, ppl5 = rref_wpp (Ab5)
show (Ab5rr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr|r} 1.00000000000000 & 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 1.00000000000000 & 1.00000000000000 & -0.000000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

ppl5

[0, 1, 4]

[0, 1, 4]

xx5r = rref_solution (Ab5rr, ppl5)
show (xx5r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,1.00000000000000,\,0.000000000000000,\,0.000000000000000,\,-1.00000000000000\right)$

Ab6rr, ppl6 = rref_wpp (Ab6)
show (Ab6rr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 0.000000000000000 & -2.00000000000000 & 0.000000000000000 & -4.44089209850063 \times 10^{-16} \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 0.000000000000000 & -3.33066907387547 \times 10^{-16} \\ 0.000000000000000 & 0.000000000000000 & -3.33066907387547 \times 10^{-16} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

show (Ab6rr.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 0.000000000000000 & -2.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & -1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

ppl6

[0, 1, 3]

[0, 1, 3]

xx6r = rref_solution (Ab6rr, ppl6)
show (xx6r)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-4.44089209850063 \times 10^{-16},\,1.00000000000000,\,0.000000000000000,\,-1.00000000000000\right)$

Ab7rr, ppl7 = rref_wpp (Ab7)
show (Ab7rr)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 0.000000000000000 & -2.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 1.00000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & -3.33066907387547 \times 10^{-16} & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & -6.66133814775094 \times 10^{-16} & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

show (Ab7rr.apply_map (chop()))

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|r} 1.00000000000000 & 0.000000000000000 & -2.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 1.00000000000000 & 2.00000000000000 & 0.000000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 & 0.000000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 1.00000000000000 \\ 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 & 0.000000000000000 \end{array}\right)$

ppl7

[0, 1, 3, 4]

[0, 1, 3, 4]

PPPK_programs_REF_and_RREF

251 days ago by fresl