@Markiyan Hirnyk 

If unknown variables are the entries of S and R, then, for example, the equation j = c1 s1 + c2 s2 + c3 s3 + c4 s4   is not resolved yet with respect to unknown variables. And it says a1=a1, c1=c1, etc. Does this mean a1, c1 are free variables?

@brian bovril 

The worksheet I used is below.

I want to use a general 4x4 matrix for T (instead of the specified 0-1 matrix you see below). Then I want to solve S.R=~T subject to the given constraints

of eq1 and eq2. I want to find what S and R must be in order to have a solution for a general T. I tried doing

this (and had the program terminate after one solution was found since I only need one, I do not need

them all), but I don't think the solution set makes sense. I want the solution of the unknowns (the entries of S and R).

@Markiyan Hirnyk 

hi there! I have a question about this please. Rather than using a specific 4x4 matrix T (as I did originally above),

I want to use a general 4x4 matrix for T. Then I want to solve S.R=~T subject to the given constraints

of eq1 and eq2 as above. I want to find what S and R must be in order to have a solution. I tried doing

this (and had the program terminate after one solution was found since I only need one, I do not need

them

all) and I got a solution. But the solution doesnt make sense since the unknown variables (the variables of

the matrices S and R) don't seem to be resolved. For example, the solution j = .... is in terms of entries of

S and R and thus doesnt seem to be resolved. I converted it to LaTeX: 

Solving the equation $T = S \cdot R$ is equivalent to solving the

polynomial system
\begin{align*}
q & = \sum_{i = 1}^4 a_i s_i, \\
w & = \sum_{i = 1}^4 a_i t_i, \\
y & = \sum_{i = 1}^4 a_i r_i, \\
u & = \sum_{i = 1}^4 a_i l_i, \\
&\\
p & = \sum_{i = 1}^4 b_i s_i, \\
f & = \sum_{i = 1}^4 b_i t_i, \\
g & = \sum_{i = 1}^4 b_i r_i, \\
h & = \sum_{i = 1}^4 b_i l_i, \\
&\\
j & = \sum_{i = 1}^4 c_i s_i, \\
k & = \sum_{i = 1}^4 c_i t_i, \\
z & = \sum_{i = 1}^4 c_i r_i, \\
x & = \sum_{i = 1}^4 c_i l_i, \\
&\\
v & = \sum_{i = 1}^4 d_i s_i, \\
n & = \sum_{i = 1}^4 d_i t_i, \\
m & = \sum_{i = 1}^4 d_i r_i, \\
o & = \sum_{i = 1}^4 d_i l_i,
\end{align*}
subject to the constraints $a_i d_i = b_i c_i$ and $s_i l_i = t_i r_i$ for $i = 1, 2, 3, 4$.
Using Maple, we determined one solution (we had the program terminate once one

solution was determined) to be:

\[
{\it a_1}={\it a_1},
\]
\[
{\it a_2}={\it a_2},
\]
\[
{\it a_3}={\it
a_3},
\]
\[
{\it a_4}={\frac {{\it b_4}\,{\it c_4}}{{\it d_4}}},
\]
\[
{\it
b_1}=0,
\]
\[
{\it b_2}=0,
\]
\[
{\it b_3}=0,
\]
\[
{\it b_4}={\it b_4},
\]
\[
{\it c_1}={
\it c_1},
\]
\[
{\it c_2}={\it c_2},
\]
\[
{\it c_3}={\it c_3},
\]
\[
{\it c_4}={\it
c_4},
\]
\[
{\it d_1}=0,
\]
\[
{\it d_2}=0,
\]
\[
{\it d_3}=0,
\]
\[
{\it d_4}={\it d_4},
\]
\[
f={
\it b_4}\,{\it t_4},
\]
\[
g={\it b_4}\,{\it r_4},
\]
\[
h={\frac {{\it b_4}\,{
\it t_4}\,{\it r_4}}{{\it s_4}}},
\]
\[
j={\it c_1}\,{\it s_1}+{\it c_2
}\,{\it s_2}+{\it c_3}\,{\it s_3}+{\it c_4}\,{\it s_4},
\]
\[
k={\it
c_1}\,{\it t_1}+{\it c_2}\,{\it t_2}+{\it c_3}\,{\it t_3}+{\it
c_4}\,{\it t_4},
\]
\[
{\it l_1}={\frac {{\it t_1}\,{\it r_1}}{{\it s_1
}}},
\]
\[
{\it l_2}={\frac {{\it t_2}\,{\it r_2}}{{\it s_2}}},
\]
\[
{\it l_3}
={\frac {{\it t_3}\,{\it r_3}}{{\it s_3}}},
\]
\[
{\it l_4}={\frac {{\it
t_4}\,{\it r_4}}{{\it s_4}}},
\]
\[
m={\it d_4}\,{\it r_4},
\]
\[
n={\it d_4}
\,{\it t_4},
\]
\[
o={\frac {{\it d_4}\,{\it t_4}\,{\it r_4}}{{\it s_4}}
},
\]
\[
p={\it b_4}\,{\it s_4},
\]
\[
q={\frac {{\it a_1}\,{\it s_1}\,{\it d_4
}+{\it a_2}\,{\it s_2}\,{\it d_4}+{\it a_3}\,{\it s_3}\,{\it d_4
}+{\it b_4}\,{\it c_4}\,{\it s_4}}{{\it d_4}}},
\]
\[
{\it r_1}={\it
r_1},
\]
\[
{\it r_2}={\it r_2},
\]
\[
{\it r_3}={\it r_3},
\]
\[
{\it r_4}={\it r_4
},
\]
\[
{\it s_1}={\it s_1},
\]
\[
{\it s_2}={\it s_2},
\]
\[
{\it s_3}={\it s_3},
\]
\[
{
\it s_4}={\it s_4},
\]
\[{\it t_1}={\it t_1},
\]
\[
{\it t_2}={\it t_2},
\]
\[
{\it
t_3}={\it t_3},
\]
\[
{\it t_4}={\it t_4},
\]
\[
u={\frac {{\it a_1}\,{\it t_1
}\,{\it r_1}\,{\it s_2}\,{\it s_3}\,{\it d_4}\,{\it s_4}+{\it
a_2}\,{\it t_2}\,{\it r_2}\,{\it s_1}\,{\it s_3}\,{\it d_4}\,{
\it s_4}+{\it a_3}\,{\it t_3}\,{\it r_3}\,{\it s_1}\,{\it s_2}\,
{\it d_4}\,{\it s_4}+{\it b_4}\,{\it c_4}\,{\it t_4}\,{\it r_4}
\,{\it s_1}\,{\it s_2}\,{\it s_3}}{{\it s_1}\,{\it s_2}\,{\it
s_3}\,{\it d_4}\,{\it s_4}}},
\]
\[
v={\it d_4}\,{\it s_4},
\]
\[
w={\frac {{
\it a_1}\,{\it t_1}\,{\it d_4}+{\it a_2}\,{\it t_2}\,{\it d_4}+{
\it a_3}\,{\it t_3}\,{\it d_4}+{\it b_4}\,{\it c_4}\,{\it t_4}}{
{\it d_4}}},
\]
\[
x={\frac {{\it c_1}\,{\it t_1}\,{\it r_1}\,{\it s_2}
\,{\it s_3}\,{\it s_4}+{\it c_2}\,{\it t_2}\,{\it r_2}\,{\it s_1
}\,{\it s_3}\,{\it s_4}+{\it c_3}\,{\it t_3}\,{\it r_3}\,{\it
s_1}\,{\it s_2}\,{\it s_4}+{\it c_4}\,{\it t_4}\,{\it r_4}\,{
\it s_1}\,{\it s_2}\,{\it s_3}}{{\it s_1}\,{\it s_2}\,{\it s_3}
\,{\it s_4}}},
\]
\[
y={\frac {{\it a_1}\,{\it r_1}\,{\it d_4}+{\it a_2}
\,{\it r_2}\,{\it d_4}+{\it a_3}\,{\it r_3}\,{\it d_4}+{\it b_4}
\,{\it c_4}\,{\it r_4}}{{\it d_4}}},
\]
\[
z={\it c_1}\,{\it r_1}+{\it
c_2}\,{\it r_2}+{\it c_3}\,{\it r_3}+{\it c_4}\,{\it r_4}
\]

@Markiyan Hirnyk 

hi there! I have a question about this please. Rather than using a specific 4x4 matrix T (as I did originally above),

I want to use a general 4x4 matrix for T. Then I want to solve S.R=~T subject to the given constraints

of eq1 and eq2 as above. I want to find what S and R must be in order to have a solution. I tried doing

this (and had the program terminate after one solution was found since I only need one, I do not need

them

all) and I got a solution. But the solution doesnt make sense since the unknown variables (the variables of

the matrices S and R) don't seem to be resolved. For example, the solution j = .... is in terms of entries of

S and R and thus doesnt seem to be resolved. I converted it to LaTeX: 

Solving the equation $T = S \cdot R$ is equivalent to solving the

polynomial system
\begin{align*}
q & = \sum_{i = 1}^4 a_i s_i, \\
w & = \sum_{i = 1}^4 a_i t_i, \\
y & = \sum_{i = 1}^4 a_i r_i, \\
u & = \sum_{i = 1}^4 a_i l_i, \\
&\\
p & = \sum_{i = 1}^4 b_i s_i, \\
f & = \sum_{i = 1}^4 b_i t_i, \\
g & = \sum_{i = 1}^4 b_i r_i, \\
h & = \sum_{i = 1}^4 b_i l_i, \\
&\\
j & = \sum_{i = 1}^4 c_i s_i, \\
k & = \sum_{i = 1}^4 c_i t_i, \\
z & = \sum_{i = 1}^4 c_i r_i, \\
x & = \sum_{i = 1}^4 c_i l_i, \\
&\\
v & = \sum_{i = 1}^4 d_i s_i, \\
n & = \sum_{i = 1}^4 d_i t_i, \\
m & = \sum_{i = 1}^4 d_i r_i, \\
o & = \sum_{i = 1}^4 d_i l_i,
\end{align*}
subject to the constraints $a_i d_i = b_i c_i$ and $s_i l_i = t_i r_i$ for $i = 1, 2, 3, 4$.
Using Maple, we determined one solution (we had the program terminate once one

solution was determined) to be:

\[
{\it a_1}={\it a_1},
\]
\[
{\it a_2}={\it a_2},
\]
\[
{\it a_3}={\it
a_3},
\]
\[
{\it a_4}={\frac {{\it b_4}\,{\it c_4}}{{\it d_4}}},
\]
\[
{\it
b_1}=0,
\]
\[
{\it b_2}=0,
\]
\[
{\it b_3}=0,
\]
\[
{\it b_4}={\it b_4},
\]
\[
{\it c_1}={
\it c_1},
\]
\[
{\it c_2}={\it c_2},
\]
\[
{\it c_3}={\it c_3},
\]
\[
{\it c_4}={\it
c_4},
\]
\[
{\it d_1}=0,
\]
\[
{\it d_2}=0,
\]
\[
{\it d_3}=0,
\]
\[
{\it d_4}={\it d_4},
\]
\[
f={
\it b_4}\,{\it t_4},
\]
\[
g={\it b_4}\,{\it r_4},
\]
\[
h={\frac {{\it b_4}\,{
\it t_4}\,{\it r_4}}{{\it s_4}}},
\]
\[
j={\it c_1}\,{\it s_1}+{\it c_2
}\,{\it s_2}+{\it c_3}\,{\it s_3}+{\it c_4}\,{\it s_4},
\]
\[
k={\it
c_1}\,{\it t_1}+{\it c_2}\,{\it t_2}+{\it c_3}\,{\it t_3}+{\it
c_4}\,{\it t_4},
\]
\[
{\it l_1}={\frac {{\it t_1}\,{\it r_1}}{{\it s_1
}}},
\]
\[
{\it l_2}={\frac {{\it t_2}\,{\it r_2}}{{\it s_2}}},
\]
\[
{\it l_3}
={\frac {{\it t_3}\,{\it r_3}}{{\it s_3}}},
\]
\[
{\it l_4}={\frac {{\it
t_4}\,{\it r_4}}{{\it s_4}}},
\]
\[
m={\it d_4}\,{\it r_4},
\]
\[
n={\it d_4}
\,{\it t_4},
\]
\[
o={\frac {{\it d_4}\,{\it t_4}\,{\it r_4}}{{\it s_4}}
},
\]
\[
p={\it b_4}\,{\it s_4},
\]
\[
q={\frac {{\it a_1}\,{\it s_1}\,{\it d_4
}+{\it a_2}\,{\it s_2}\,{\it d_4}+{\it a_3}\,{\it s_3}\,{\it d_4
}+{\it b_4}\,{\it c_4}\,{\it s_4}}{{\it d_4}}},
\]
\[
{\it r_1}={\it
r_1},
\]
\[
{\it r_2}={\it r_2},
\]
\[
{\it r_3}={\it r_3},
\]
\[
{\it r_4}={\it r_4
},
\]
\[
{\it s_1}={\it s_1},
\]
\[
{\it s_2}={\it s_2},
\]
\[
{\it s_3}={\it s_3},
\]
\[
{
\it s_4}={\it s_4},
\]
\[{\it t_1}={\it t_1},
\]
\[
{\it t_2}={\it t_2},
\]
\[
{\it
t_3}={\it t_3},
\]
\[
{\it t_4}={\it t_4},
\]
\[
u={\frac {{\it a_1}\,{\it t_1
}\,{\it r_1}\,{\it s_2}\,{\it s_3}\,{\it d_4}\,{\it s_4}+{\it
a_2}\,{\it t_2}\,{\it r_2}\,{\it s_1}\,{\it s_3}\,{\it d_4}\,{
\it s_4}+{\it a_3}\,{\it t_3}\,{\it r_3}\,{\it s_1}\,{\it s_2}\,
{\it d_4}\,{\it s_4}+{\it b_4}\,{\it c_4}\,{\it t_4}\,{\it r_4}
\,{\it s_1}\,{\it s_2}\,{\it s_3}}{{\it s_1}\,{\it s_2}\,{\it
s_3}\,{\it d_4}\,{\it s_4}}},
\]
\[
v={\it d_4}\,{\it s_4},
\]
\[
w={\frac {{
\it a_1}\,{\it t_1}\,{\it d_4}+{\it a_2}\,{\it t_2}\,{\it d_4}+{
\it a_3}\,{\it t_3}\,{\it d_4}+{\it b_4}\,{\it c_4}\,{\it t_4}}{
{\it d_4}}},
\]
\[
x={\frac {{\it c_1}\,{\it t_1}\,{\it r_1}\,{\it s_2}
\,{\it s_3}\,{\it s_4}+{\it c_2}\,{\it t_2}\,{\it r_2}\,{\it s_1
}\,{\it s_3}\,{\it s_4}+{\it c_3}\,{\it t_3}\,{\it r_3}\,{\it
s_1}\,{\it s_2}\,{\it s_4}+{\it c_4}\,{\it t_4}\,{\it r_4}\,{
\it s_1}\,{\it s_2}\,{\it s_3}}{{\it s_1}\,{\it s_2}\,{\it s_3}
\,{\it s_4}}},
\]
\[
y={\frac {{\it a_1}\,{\it r_1}\,{\it d_4}+{\it a_2}
\,{\it r_2}\,{\it d_4}+{\it a_3}\,{\it r_3}\,{\it d_4}+{\it b_4}
\,{\it c_4}\,{\it r_4}}{{\it d_4}}},
\]
\[
z={\it c_1}\,{\it r_1}+{\it
c_2}\,{\it r_2}+{\it c_3}\,{\it r_3}+{\it c_4}\,{\it r_4}
\]

Thanks! When I tried it for the matrix T such that the entries were the first 16 primes, I got solutions containing _Z and _Z^2. What does this mean?

Thanks! When I tried it for the matrix T such that the entries were the first 16 primes, I got solutions containing _Z and _Z^2. What does this mean?

Yeah I think I don't need to do it for an arbitrary T yet... I'm confident that I can find a specific T such that the equation does not hold. All I need is one counter-example.

Yeah I think I don't need to do it for an arbitrary T yet... I'm confident that I can find a specific T such that the equation does not hold. All I need is one counter-example.

Thank you very very much!

Thank you very very much!

Hmm, actually all of the matrices will be 2x4 that need to be reduced, so as you said it might be convenient to just hard code it. 

Hmm, actually all of the matrices will be 2x4 that need to be reduced, so as you said it might be convenient to just hard code it. 

Thanks very much! 

Thanks very much!