Is a diagonal matrix with "$12$"s along the diagonal an identity matrix? Is C following an identity matrix?
$$A = \begin{bmatrix}
 2& -1 & 12 \\ 
 3 & 6 & -9\\ 
 1& 1& 3  
\end{bmatrix}$$
$$B = \begin{bmatrix}
 9 & 5 & -21 \\ 
 -6 & -2 & 18\\ 
 -1& -1& 5  
\end{bmatrix}$$
$$C = \begin{bmatrix}
 12& 0 & 0 \\ 
 0 & 12 & 0\\ 
 0& 0&12  
\end{bmatrix}$$
I perform AB and BA and end up with C as the answer. I do know that an identity matrix must consist of only ones and zeros. All the diagonals are 12 and if you pull out 1/12, the identity would remain. Would it alone be considered the identity due the numbers being the same? I’m guessing no but am not sure.

In this situation, $C$ is a scalar matrix that is, as you say, $12$ times the identity matrix, i.e. $C = 12I$. This is not the same thing as the identity, but it has some special properties in common with the identity matrix. For example, it commutes with any other $3 \times 3$-matrix, i.e. $CX = XC$ for any $X$.
Also, the fact that $AB = BA = C$ allows you write down the inverse of either $A$ or $B$ with ease:
$$
AB = C 
\implies 
A\, \bigl( \tfrac{1}{12} B \bigr) = \tfrac{1}{12} C = I, 
$$
which shows that
$$
A^{-1} = \tfrac{1}{12} B 
= \tfrac{1}{12} \begin{bmatrix}
 9 &  5 & -21 \\ 
-6 & -2 &  18 \\ 
-1 & -1 &   5  
\end{bmatrix}
= \begin{bmatrix}
 \tfrac{3}{4}  &  \tfrac{5}{12} & -\tfrac{7}{4}  \\ 
-\tfrac{1}{2}  & -\tfrac{1}{6}  &  \tfrac{3}{2}  \\ 
-\tfrac{1}{12} & -\tfrac{1}{12} &  \tfrac{5}{12}  
\end{bmatrix}.
$$
