Determining matrix $A$ and $B$, rectangular matrix

Let $A$ be a $3\times 2$ matrix and $B$ be a $2\times 3$ be matrices satisfying $$AB=\begin{pmatrix} 8 & 2 & -2\\ 2 & 5 & 4\\ -2 & 4 & 5\end{pmatrix}$$ Calculate $BA$. How would you go for this problem? Do we start by noticing the matrix is symmetric? Any hints/ideas? Thanks

This is a strange problem. In general, the quesiton would not be well posed. As observed in the comment by @JohnMa it is neccesary that the given $3\times 3$-matrix has rank two. This is satisfied, but even then, the question would not be well posed in gerneral. Namely, if $A$ and $B$ are such that $AB$ is your given matrix, then for any invertible $2\times 2$ matrix $T$ also $AT$ and $T^{-1}B$ have the same property. In that way $BA$ gets replaced by $T^{-1}BAT$, which is different from $BA$ in general. Hence the only case in which the question is well posed is if $BA$ is a multiple of the identity matrix, and with the given $3\times 3$-matrix, you indeed get $BA=9I$.

EDIT: I have rewritten and extended the next paragraph a bit, following the remark by @CarlosMendoza:

The way to prove this, is as follows. You first compute the eigenvalues of $AB$ and find that they are $0$ (multiplicity one) and $9$ (multiplicity two). In particular $AB$ has rank two, and viewing it as a linear map $\mathbb R^3\to\mathbb R^3$ its kernel is given by the $0$-eigenspace and its image is given by the $9$-eigenspace. (The first equality holds by definition, while for the second, the eigenspace is obviously contained in the image and has the same dimension.) Now the matrices $A$ and $B$ correspond to linear maps $\mathbb R^2\to\mathbb R^3$ and $\mathbb R^3\to \mathbb R^2$, respectively, so they both have rank at most two. By definition, the kernel of $B$ is contained in the kernel of $AB$ and the image of $AB$ is contained in the image of $A$. Since the kernel of $B$ has at least dimension one and the image of $A$ has at most dimension two, both these inclusions must be equalities. In particular, the image of $A$ is the $9$--eigenspace of $AB$, and calling this $V$, $A$ must define a linear isomorphism $\mathbb R^2\to V$. But this immediately implies that $(AB)A=9A$ since $AB$ is multiplication by $9$ on $V$. Rewriting the left hand side as $A(BA)$, you get $A(BAx)=A(9x)$ for all $x\in\mathbb R^2$, and since $A$ is injective, this implies $BAx=9x$ for all $x\in\mathbb R^2$.

I'm sorry I don't have time to develop al the exercise, but I think I can give you some hints. After that I think is more or less straight-forward. Consider \begin{equation} A=\left(\begin{array}{ccc} a & c & e\\ b & d & f\\ 0 & 0 & 0 \end{array}\right) \end{equation} \begin{equation} B=\left(\begin{array}{ccc} g & h & 0\\ i & l & 0\\ m & n & 0 \end{array}\right) \end{equation} So we have \begin{equation} BA=\left(\begin{array}{ccc} u & v & 0\\ w & z & 0\\ 0 & 0 & 0 \end{array}\right) \end{equation} Some invariants are: \begin{equation} trAB=trBA \end{equation} And from this we have
\begin{equation} u+z=18 \end{equation} Since AB is simmetric so is BA \begin{equation} v=w \end{equation} The Binet formula shouldn't be so useful because detAB=detBA=0 Anyway I think that from here more or less You can proceed with direct calculation.Let me know.

Using the suggestion of @WillJagy and the direct approach suggested by @Dac0, we have $$ A = \begin{bmatrix} a & b\\ c & d\\ e & f\\ \end{bmatrix}\qquad B = \begin{bmatrix} a & c & e\\ b & d & f\\ \end{bmatrix} $$

and from $AB$ we get the following set of equations

\begin{align} a^2 + b^2 &= 8\\ c^2 + d^2 &= 5\\ e^2 + f^2 &= 5\\ ac + bd &= 2\\ ae + fb &= -2\\ ec + fd &= 4\\ \end{align}

I solved this in WolframAlpha and tested with this particular solution(out of infinite solutions):

$$ A = \begin{bmatrix} -2 & -2\\ -2 & 1\\ -1 & 2\\ \end{bmatrix}\qquad B = \begin{bmatrix} -2 & -2 & -1\\ -2 & 1 & 2\\ \end{bmatrix} $$

Interestingly enough,

$$BA = \begin{bmatrix} 9 & 0\\ 0 & 9\\ \end{bmatrix} $$

which corresponds to the solution proposed by @AndreasCap. It's also interesting to note that the columns of $A$ are orthogonal and their squared length are equal to $9$, the non-zero eigenvalue of $AB$ with multiplicity $2$.

I gave importance to @WillJagy suggestion because somehow it remembered me of the covariance matrix. I am not saying is the same, but there are nice similarities.