How to solve this six variables of simultaneous equation?

Question

The problems is:

$A^2+D^2=5$

$B^2+E^2=2$

$C^2+F^2=6$

$AB+DE=3$

$BC+EF=1$

$AC+DF=1$

Please tell me what is the roots for this simultaneous equation, no process or just solve it by calculator are both fine to me.

Thank you.

Answer 1

You can reduce to three unknowns $x,y,z$ making $$A=\sqrt5\sin(x),\space D=\sqrt5\cos(x)\\B=\sqrt2\sin(y),\space E=\sqrt2\cos(y)\\C=\sqrt6\sin(z),\space F=\sqrt6\cos(z)\\AB+DE=\sqrt{10}\cos(x-y)=3\\BC+EF=\sqrt{12}\cos(y-z)=1\\AC+DF=\sqrt{30}\cos(x-z)=1$$ This gives $$\arccos\left(\dfrac{3}{\sqrt{10}}\right)=x-y\approx0.3217505\\\arccos\left(\dfrac{1}{\sqrt{12}}\right)=y-z\approx1.2779535\\\arccos\left(\dfrac{1}{\sqrt{30}}\right)=z-x\approx1.3871923$$ So you can verify that there is not solution to your system.

Answer 2

I obtain, for the RHS equal to $2$, $$ A=\sqrt{2 - F^2}, \; B=\sqrt{2-F^2},\; C=\sqrt{2- F^2},\; D=E=F, $$ or the negative square roots. So $A=B=C$ and $D=E=F$.

The edited problem has no solution (for RHS $5,2,6,3,1,1$).

Answer 3

There is no triple of (real) vectors in $\mathbb R^2$ with the desired Gram matrix; the angles

There are triples of real vectors in $\mathbb R^3$ it can be done, not with integers. The quick proof is simply that the Gram matrix has determinant $5,$ not a square number.

Finally, this arrangement exists with integer coordinates in $\mathbb R^4,$ the basis vectors are the rows of $$ B = \left( \begin{array}{cccc} 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 2 \\ \end{array} \right) $$

$$ G = BB^T = \left( \begin{array}{ccc} 5 & 3 & 1 \\ 3 & 2 & 1 \\ 1 & 1 & 6 \\ \end{array} \right) $$