Vtyjkyuk

If $K$ is a field and $A= beginpmatrix a & 0 &0 \ 0 & b &0 \ 0& 0& c endpmatrix$ , $B=beginpmatrix 0 & p &0 \ 0 & 0 &q \ r& 0& 0 endpmatrix$ , $C=beginpmatrix 0 & 0 &x \ y & 0 &0 \ 0& z& 0 endpmatrix$ ,

where $a,b,c,p,q,r,x,y,zin K$ are such that $S:=a,b,c,p,q,r,x,y,z$ is closed under multiplication , then I can show that $A^2,B^2,C^2,AB,BA,BC,CB,AC,CA in Bigg beginpmatrix a' & 0 &0 \ 0 & b' &0 \ 0& 0& c' endpmatrix, beginpmatrix 0 & p' &0 \ 0 & 0 &q' \ r'& 0& 0 endpmatrix, beginpmatrix 0 & 0 &x' \ y' & 0 &0 \ 0& z'& 0 endpmatrix : a',b',c',p',q',r',x',y',z' in S Bigg$

My question is: Can we use this fact to construct a non-abelian group of order $27$ ?

The only way I know of constructing a non-abelian group of order $p^3$ for odd prime $p$ is using upper triangular $3times3$ matrices or special kind of $2times 2$ matrices with entries in $mathbb Z/(p)$ or $mathbb Z/(p^2)$, as given here http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf .

Please help.

Thanks in advance

Yes, provided that $K$ contains a primitive cube root $omega$ of $1$. The two nonabelian groups of order $27$ both have irreducible representations of degree $3$ that are induced from $1$-dimensional representations of elementary abelian subgroups of order $9$.

The nonabelian group of exponent $3$ $is$
$$leftlangle,left(beginarrayccc1&0&0 \ 0&omega&0 \ 0&0&omega^2 endarrayright),,
left(beginarrayccc0&0&1\1&0&0\0&1&0endarrayright)
,rightrangle$$
and the one of exponent $9$ is
$$leftlangle,left(beginarrayccc1&0&0 \ 0&omega&0 \ 0&0&omega^2 endarrayright),,
left(beginarrayccc0&0&omega\1&0&0\0&1&0endarrayright)
,rightrangle.$$
To see that the first group has order $27$, call the two generators $a$ and $b$. Then $b$ is a permutation matrix, and $b^-1ab = left(beginarraycccomega&0&0\0&omega^2&0\0&0&1endarrayright)$, so $b^-1aba^-1$ is the scalar matrix $omega I_3$, which commutes with $a$ and $b$.
So, putting $c=b^-1aba^-1$, $a$, $b$ and $c$ satisfy the relations of the presentation $$langle a,b,c mid a^3=b^3=c^3=1, c=b^-1aba^-1, [a,c]=[b,c]=1rangle,$$
which define a group of order $27$. It's not hard to see that the group generated by $a$ and $b$ has order greater than $9$, so it must have order $27$.
You can prove that the second group has order $27$ with a similar argument.