Solution to matrix equation Say I have row vector $x=[a,b,c]$ with length 3, the 3 by 3 matrix $M=\pmatrix{M_{11} & M_{12} & M_{13} \cr M_{21} & M_{22} & M_{23} \cr M_{31} & M_{32} & M_{33} \cr}$ and 3 by 3 matrix $X=x^T*x$) which combined give vector $y=[abc,bc,c]$. How do I solve this equation? $xMX=y$
or
$$[a,b,c]\pmatrix{M_{11} & M_{12} & M_{13} \cr M_{21} & M_{22} & M_{23} \cr M_{31} & M_{32} & M_{33} \cr}\pmatrix{aa & ab & ac \cr ab & bb & cb \cr ac & bc & cc \cr}=[abc,bc,c]$$
or
$$[aM_{11}+bM_{12}+cM_{13},aM_{21}+bM_{22}+cM_{23},aM_{31}+bM_{32}+cM_{33}]\pmatrix{aa & ab & ac \cr ab & bb & cb \cr ac & bc & cc \cr}=[abc,bc,c]$$
My actual goal is to go from $x$ to $y$ through linear algebra. We came up with this matrix multiplication. Now I want to solve for M and I am not sure how to approach this. Any help greatly appreciated.

Find xMX.
You'll get 9 terms in every column.
Among 9 terms of first column, coefficient of term containing abc will be 1(because in result,only abc is there, & its coef is 1), while all other coefficients will be zero. 
For 2nd column, coef of bc will be 1, rest all coef will be zero, & for 3rd column, coef of c will be 1, rest all coef will be zero.
So, in total, you'll get 27 equations. Solve them & you'll get all 9 values of M.
