In my work (this has some practical application I won't get into), I ended up with a system of equations like this:
\(\displaystyle \displaystyle A_1 = x_1 + y_1 \)
\(\displaystyle \displaystyle A_2 = x_2 + y_2 \)
\(\displaystyle \displaystyle B_1 = x_1 + a_1y_1 \)
\(\displaystyle \displaystyle B_2 = x_2 + a_1y_2 \)
\(\displaystyle \displaystyle C_1 = x_1 + a_2y_1 \)
\(\displaystyle \displaystyle C_2 = x_2 + a_2y_2 \)
where \(\displaystyle \{A_1, A_2, B_1, B_2, C_1, C_2\}\) are known quantities, and \(\displaystyle \{x_1, y_1, x_2, y_2, a_1, a_2\} \) are unknown quantities that I would like to solve for. So, the system has 6 equations and 6 unknowns. If I go through and solve the system using substitution/elimination, I end up with equations for each of the variables in terms of the others (equations I have verified numerically):
\(\displaystyle \displaystyle y_2 = \frac{C_2 - B_2}{a_2 - a_1} \)
\(\displaystyle \displaystyle x_2 = C_2 - a_2\frac{C_2 - B_2}{a_2 - a_1} \)
\(\displaystyle \displaystyle y_1 = \frac{C_1 - B_1}{a_2 - a_1} \)
\(\displaystyle \displaystyle x_1 = C_1 - a_2\frac{C_1 - B_1}{a_2 - a_1} \)
and
\(\displaystyle \displaystyle a_1 = a_2\left(\frac{W+1}{W}\right) - \frac{1}{W} \)
where
\(\displaystyle \displaystyle W \equiv \frac{A_2 - C_2}{C_2-B_2} \)
But I've found out that you cannot solve for the last unknown, \(\displaystyle a_2\), because you end up with
\(\displaystyle \displaystyle a_2 = \frac{Va_1 + 1}{V+1}\)
where
\(\displaystyle \displaystyle V \equiv \frac{A_1 - C_1}{C_1-B_1} \)
Unfortunately, this leads to the equation
\(\displaystyle \displaystyle \left(1-\frac{V}{W}\right)a_2 = 1-\frac{V}{W} \)
You can show that \(\displaystyle V = W \) and hence we end up with
\(\displaystyle \displaystyle 0\cdot a_2 = 0 \)
I guess my question is, is there some way to understand, just by looking at the original system, that it will be degenerate, or to highlight what property of the original set of equations is leading to the degeneracy?
Thanks.
\(\displaystyle \displaystyle A_1 = x_1 + y_1 \)
\(\displaystyle \displaystyle A_2 = x_2 + y_2 \)
\(\displaystyle \displaystyle B_1 = x_1 + a_1y_1 \)
\(\displaystyle \displaystyle B_2 = x_2 + a_1y_2 \)
\(\displaystyle \displaystyle C_1 = x_1 + a_2y_1 \)
\(\displaystyle \displaystyle C_2 = x_2 + a_2y_2 \)
where \(\displaystyle \{A_1, A_2, B_1, B_2, C_1, C_2\}\) are known quantities, and \(\displaystyle \{x_1, y_1, x_2, y_2, a_1, a_2\} \) are unknown quantities that I would like to solve for. So, the system has 6 equations and 6 unknowns. If I go through and solve the system using substitution/elimination, I end up with equations for each of the variables in terms of the others (equations I have verified numerically):
\(\displaystyle \displaystyle y_2 = \frac{C_2 - B_2}{a_2 - a_1} \)
\(\displaystyle \displaystyle x_2 = C_2 - a_2\frac{C_2 - B_2}{a_2 - a_1} \)
\(\displaystyle \displaystyle y_1 = \frac{C_1 - B_1}{a_2 - a_1} \)
\(\displaystyle \displaystyle x_1 = C_1 - a_2\frac{C_1 - B_1}{a_2 - a_1} \)
and
\(\displaystyle \displaystyle a_1 = a_2\left(\frac{W+1}{W}\right) - \frac{1}{W} \)
where
\(\displaystyle \displaystyle W \equiv \frac{A_2 - C_2}{C_2-B_2} \)
But I've found out that you cannot solve for the last unknown, \(\displaystyle a_2\), because you end up with
\(\displaystyle \displaystyle a_2 = \frac{Va_1 + 1}{V+1}\)
where
\(\displaystyle \displaystyle V \equiv \frac{A_1 - C_1}{C_1-B_1} \)
Unfortunately, this leads to the equation
\(\displaystyle \displaystyle \left(1-\frac{V}{W}\right)a_2 = 1-\frac{V}{W} \)
You can show that \(\displaystyle V = W \) and hence we end up with
\(\displaystyle \displaystyle 0\cdot a_2 = 0 \)
I guess my question is, is there some way to understand, just by looking at the original system, that it will be degenerate, or to highlight what property of the original set of equations is leading to the degeneracy?
Thanks.