<s>[INST] Matrix multiplication computation Any tips how to solve this?
$$
\left[ \begin{matrix}1 & 2 & 0 \\ -2 & -5 & 1 \\ 11&15&5 \end{matrix}\right] \times \mathbf{X} \times \left[ \begin{matrix} -4&5&1\\ -4&5&1 \\ -4&5&1\end{matrix}\right] =\left[ \begin{matrix}-12 & 15&3 0 \\ -24&-30&6 \\ -125&150&31  \end{matrix}\right]
$$
 [/INST] First instinct would be to multiply on the left and right by inverses to get $A$ on its own on the LHS. However the rightmost of these matrices is not invertible. But we can get 
$$
A \times \left[\begin{array}{ccc}-4 & 5 & 1\\-4 & 5 & 1\\-4 & 5 & 1 \end{array}\right] = \left[\begin{array}{ccc} 235 & 0 & -599 \\ -\frac{247}{2} & \frac{15}{2} & \frac{629}{2} \\ -\frac{343}{2} & \frac{15}{2} & \frac{761}{2}\end{array}\right]
$$
The system of equations for the components of $A$ is over-determined. This means that, letting
$$
A = \left[ \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]
$$
we see (for instance) that this requires
$$
-4 (a_1 + a_2 + a_3) = 235 \\
\implies a_1 + a_2 + a_3 = -\frac{235}{4}
$$
but 
$$
5 (a_1 + a_2 + a_3) = 0 \\
\implies a_1 + a_2 + a_3 = 0
$$
which is clearly contradictory. So no solution exists.
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