I am in pickle to figure out a way to input this somewhat odd system of simultaneous equations in Mathematica
The system reads
$$a_1+2 a_2-2 a_3=3 a_2-a_3=-a_6=-a_7=a_2-a_3$$ $$a_1+a_2-a_3-a_6=2 a_2-a_6=4 a_2-2 a_3=-a_6$$ $$a_1+a_2-a_3-a_7=2 a_2-a_7$$
The solution to this peculiar system is $a_2=\frac{1}{4}a_1=\frac{1}{3}a_3$, $a_6=a_7=0$.
Here is my try
Clear[a1, a2, a3, a6, a7];
Eq1=a1+2 a2-2 a3==3 a2-a3
Eq2=a1+2 a2-2 a3==-a6
Eq3=a1+2 a2-2 a3==-a7
Eq4=a1+2 a2-2 a3==a2-a3
Eq5=a1+a2-a3-a6==2 a2-a6
Eq6=a1+a2-a3-a6==4 a2-2 a3
Eq7=a1+a2-a3-a6==-a6
Eq8=a1+a2-a3-a7==2 a2-a7
Solve[{Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, Eq7, Eq8}, {a1, a2, a3, a6,a7}]
gives me a trivial solution.
Is there a way around?
{Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, Eq7, Eq8} /. {a1 -> 4 a2, a3 -> 3 a2, a6 -> 0, a7 -> 0}
return ? $\endgroup$Last[CoefficientArrays[List @@ FullSimplify[And @@ {a[1] + 2 a[2] - 2 a[3] == 3 a[2] - a[3] == -a[6] == -a[7] == a[2] - a[3], a[1] + a[2] - a[3] - a[6] == 2 a[2] - a[6] == 4 a[2] - 2 a[3] == -a[6], a[1] + a[2] - a[3] - a[7] == 2 a[2] - a[7]}], {a[1], a[2], a[3], a[6], a[7]}]]
yields a nonsingular coefficient matrix (hence, no null space, and only the all-zeros solution is possible). $\endgroup$