Supposing the following linear system of equations has always a single solution: $$ a_1X + b_1Y +c_1Z = d_1 $$ $$ a_2X + b_2Y +c_2Z = d_2 $$ $$ a_3X + b_3Y +c_3Z = d_3 $$
Where $a_i,b_i,c_i$ are measurements from sensors with a specific resolution which means that they are not 100% accurate and their real values may differ with an error $e \in [-0.5, +0.5]$
By solving the system using those non accurate $a_i,b_i,c_i, d_i$, I will got $X, Y, Z$ but with tolerance.
I noticed that I can propose another equations that contains $X, Y, Z$ which is: $$ X^2 + Y^2 + Z^2 = r^2 $$ Where the actual value of $r$ is known without any error.
How can I use this 4th equation to decrease the error in computing $X, Y, Z$ from the original system of equations?
Is there a mathematical name for this technique (decreasing error by adding another constraint or equation?
Extending the question: Is it possible to increase the accuracy if $r$ is not 100% accurate with the same error range of the other values?
If we have more equations than variables we could try solve $$ \min_{x} ||Ax - b||^2 $$ This has the solution $$ x = (A^tA)^{-1}A^tb $$ https://en.wikipedia.org/wiki/Least_square. So using this method you can try using the extra information. Otherwise independent extra measurements will increase the precition in the result. For your case in particular you could use first skip it and get a preliminary solution vector $X_1$, now solve for $X_2$ with $X_1'X_2 = r^2$ added and repeat. until (hopefully) convergence. Try it to see if it improves the precition. For your specified problem $(b=0)$ you don't have a unique solution untill you use your constraint which will fix it in case the kernel of $A$ has dimension 1. To find out first use single value decomposition of $A$. Because the inexactness of $A$ and the dimention of the kernel is 0 you can examin the diagonal matrix to see if there exist an element close to zero and just take the corresponding vector associated to that diagonal entry and scale that vector so that the norm of it is $r$.