Solving a System of Six unknowns using Six Equations As the title states, the problem seems fairly simple, but I'm having great difficulties with it. I am solving for a matrix which consists of three vectors made up of nine unknowns (we can cut it down to six using known symmetries). So, six unknowns, six equations that I have derived. Nothing too tricky. I'm going to name the unknowns: a, e, i, b, d, and c. The equations are as follows:
$a^2 + e^2 + i^2 = 29.1$
$b^2 + e^2 + d^2 = 29.1$
$c^2 + d^2 + i^2 = 156.25$
$(ae)+(be)+(di) = -14.55$
$(bd)+(cd)+(ei) = 0$
$(de)+(ai)+(ci) = 0$
That's literally it. I've spent a lot of time messing around with these equations but I've gotten nowhere, so maybe there's something that I'm missing. Any help at all would be appreciated.
edit: Additional background information. The matrix consists of (a1, a2, a3; b1, b2, b3; c1, c2, c3). Symmetries: a2=b1, a3=c1, b3=c2. The magnitude of vector a=5.394, b=5.394, c=12.5. Cos of vectors 12 = -.5. Cos of vectors 13, and 23 =0. This is the entirety of the information that I have to work with.
 A: Note that if we form the vectors $\vec{v}_1=\langle a, e, i\rangle$, $\vec{v}_2=\langle e, b, d\rangle$ and $\vec{v}_3=\langle i, d, c\rangle$ then the six written equations can be written as equations for the six dot products among these vectors (the three norms $|\vec{v}_i|^2$ and the three dot products $\vec{v}_i\cdot\vec{v}_j$ for $i\neq j$). In particular this means that the whole equation can be written as $\mathbf{M}^2=\mathbf{A}$ for the $3\times 3$ matrix $\mathbf{M}$ made up of the three vectors and a suitable constant symmetric matrix $\mathbf{A}$.
This can generally be solved by diagonalizing the matrix $\mathbf{A}$; if $\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}$ for diagonal $\mathbf{D}$ and $\mathbf{M}=\mathbf{P}\mathbf{M'}\mathbf{P}^{-1}$, then it's easy to see that $\mathbf{M'}^2=\mathbf{D}$, but this last equation can easily solved. The flexibility allowed by reflections (i.e., matrices $\mathbf{R}$ such that $\mathbf{R}^2=\mathbf{I}$) can then be used to try and force the equalities among coefficients that you're looking for.  (This is one reason to potentially work with the original set of variables, because it may be possible to write these scalar equality constraints in terms of more easily-solved vector constraints.) (ETA: And note that the 'original' version of the problem can be similarly specified in terms of computing a matrix square root of a symmetric matrix; this is a well-studied problem, and you should be able to find more information under that name.)
A: Using Maple's Groebner basis package, I get the following result . . .

A $6$-tuple $(a,b,c,d,e,i)$ satisfies the system if and only if
\begin{align*}
&d = i = 0\\[6pt]
&c = {\small{\pm \frac{25}{2}}}\\[6pt]
&a = \pm \sqrt{{\small{\frac{291}{20}}}\left(1 \pm {\small{\frac{1}{2}}}\sqrt{3}\right)}\\
&b = a\\[6pt]
&e = {\small{\frac{40}{291}}}a^3-4a
\end{align*}
A: You say you haven't had a full answer with Matlab.
In fact, it is possible to use Matlab to get all unknowns.
Here is the script (I have chosen to replace $i$ by $f$, in order to avoid problems with $i$ considered as a complex number):
  syms a b c d e f
  [a,b,c,d,e,f]=solve(a^2+e^2+f^2==29.1,...
      b^2+e^2+d^2==29.1,...
      c^2+d^2+f^2==156.25,...
      (a*e)+(b*e)+(d*f)==-14.55,...
      (b*d)+(c*d)+(e*f)==0,...
      (d*e)+(a*f)+(c*f)==0)
  simplify(a)
  simplify(b)
  simplify(c)
  simplify(d)
  simplify(e)
  simplify(f)

The answers are the same as @quasi has found but under the form:


*

*$a=b=\dfrac{1}{20}\left(\sqrt{1455}-3\sqrt{485}\right)$

*$c=\pm\dfrac{25}{2}$

*$d=f=0$

*$e=-\dfrac{1}{20}\left(\sqrt{1455}+3\sqrt{485}\right)$
A little remark: for $a$ there is a slight difference; numerical conversion shows that:
$a=\pm\sqrt{\dfrac{291}{20}\left(1-\dfrac{\sqrt{3}}{2}\right)}$
(no $\pm$ sign inside the square root).
