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Let's compute some solution.

We will fix B as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$ and satisfy the second equation.

We will introduce $n^2$ variables for elements of $n$ and one more to ensure that $\det A\ne 0$. Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the first equation, and while it's positive, try to make some random variable assignment.

Here is a sample code:

import random
def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    # construct list of equations
    E = []
    E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
    E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

    while (d := (J := R.ideal(E)).dimension()):
        # print('Dim =',d)
        if d < 0: E.pop()    # remove last assignment
        E.append( random.choice(x) - ZZ.random_element() )

    V = J.variety()[0]     # any solution
    return A.map_apply(V.__getitem__)

Calling find_A(3) can produce a matrix like

[ 16  -1   0]
[273   1   0]
[  0  -1 -17]

Let's compute some solution.

We will fix B $B$ as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$ and satisfy the second equation.

We will introduce $n^2$ variables for elements of $n$ $A$, and one more to ensure that $\det A\ne 0$. $(\det A)^{-1}$ (to make sure that the determinant is invertible). Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the first equation, and while it's positive, try to make some random variable assignment.assignment until it drops to 0 and enable application of .variety() method.

Here is a sample code:

import random
def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    # construct list of equations
    E = []
    E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
    E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

    while (d := (J := R.ideal(E)).dimension()):
        # print('Dim =',d)
        if d < 0: E.pop()    # remove last assignment
        E.append( random.choice(x) - ZZ.random_element() )

    V = J.variety()[0]     # any solution
    return A.map_apply(V.__getitem__)

Calling find_A(3) can produce a matrix like

[ 16  -1   0]
[273   1   0]
[  0  -1 -17]

Let's compute some solution.

We will fix $B$ as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$ and satisfy the second equation.

We will introduce $n^2$ variables for elements of $A$, and one more to ensure that $(\det A)^{-1}$ (to make sure that the determinant is invertible). Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the first equation, and while it's positive, try to make some random variable assignment until it drops to 0 and enable application of .variety() method.

Here is a sample code:

import random
def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    # construct list of equations
    E = []
    E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
    E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

    while (d := (J := R.ideal(E)).dimension()):
        # print('Dim =',d)
        if d < 0: E.pop()    # remove last assignment
        E.append( random.choice(x) - ZZ.random_element() )

    V = J.variety()[0]     # any solution
    return A.map_apply(V.__getitem__)
A.apply_map(V.__getitem__)

Calling find_A(3) can produce a matrix like

[ 16  -1   0]
[273   1   0]
[  0  -1 -17]

Let's compute some solution.

We will fix $B$ as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$ and satisfy the second equation.

We will introduce $n^2$ variables for elements of $A$, and one more to ensure that $(\det A)^{-1}$ (to make sure that the determinant is invertible). Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the first equation, and while it's positive, try to make some random variable assignment until it drops to 0 and enable application of .variety() method.

Here is a sample code:

import random
 def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    while True:
        # construct list of equations
     E = []
     E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
     E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

     while (d := (J := R.ideal(E)).dimension()):
         # print('Dim =',d)
         if d < 0: E.pop()    # remove last assignment
         E.append( random.choice(x) - ZZ.random_element() )

     V = J.variety()[0]     # any solution
    J.variety()
        if V: return A.apply_map(V.__getitem__)
A.apply_map(V[0].__getitem__)

Calling find_A(3) can produce a matrix like

[ 16  -1   0]
[273   1   0]
[  0  -1 -17]

UPDATED 2025-02-16. Let's compute some solution.

We will fix $B$ as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$ and satisfy the second equation. $B^2=0$.

We will introduce $n^2$ variables for elements of $A$, and one more to ensure that $(\det A)^{-1}$ (to make sure that the determinant is invertible). Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the first equation, two equations, and while it's positive, try to make some random variable assignment until it drops to 0 and enable application of .variety() method.

Here is a sample code:

import random

def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    while True:
        # construct list of equations
        E = []
        E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
        E.extend( (B*A*B).list() )                        # second equation
        E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

        while (d := (J := R.ideal(E)).dimension()):
            # print('Dim =',d)
            if d < 0: E.pop()    # remove last assignment
            E.append( random.choice(x) - ZZ.random_element() )

        V = J.variety()
        if V: return A.apply_map(V[0].__getitem__)

Calling find_A(3) can produce a matrix like

[ 16  -1   0]
[273   1  [-1  0  0]
[ 7  0  1]
[ 5 -1 -17]
 1]

UPDATED 2025-02-16. Let's compute some solution.

We will fix $B$ as a matrix with just one off-diagonal nonzero entry, say, $B_{n,1}=1$. This will immediately imply $B^2=0$.

We will introduce $n^2$ variables for elements of $A$, and one more to ensure that $(\det A)^{-1}$ (to make sure that the determinant is invertible). Then we will compute dimension of polynomials corresponds to the entries of the left-hand side of the two equations, and while it's positive, try to make some random variable assignment until it drops to 0 and enable application of .variety() method.

Here is a sample code:

import random

def find_A(n):
    R = PolynomialRing(QQ, n^2+1, 'x')
    x = R.gens()[:n^2]

    A = Matrix(R, n, n, x)
    B = Matrix(R, n, n)
    B[n-1,0] = 1

    while True:
        # construct list of equations
        E = []
        E.extend( (A^2*B + A*B*A + B*A^2).list() )   # first equation
        E.extend( (B*A*B).list() )                        # second equation
        E.append( A.det()*R.gens()[-1] - 1 )             # det(A) != 0

        while (d := (J := R.ideal(E)).dimension()):
            # print('Dim =',d)
            if d < 0: E.pop()    # remove last assignment
            E.append( random.choice(x) choice(x) - ZZ.random_element() )

        V = J.variety()
        if V: return A.apply_map(V[0].__getitem__)

Calling find_A(3) can produce a matrix like

[-1  0  0]
[ 7  0  1]
[ 5 -1  1]