How to proceed when a Gaussian Elimination gives a row of all zeroes? I'm trying to solve a system of equations using Gaussian Elimnation, but I'm not sure how to proceed. Here are my steps:

The starting matrix (apologies for the verbose coefficients):

$$\left(\begin{array}{ccc|c}
1.732050808 &amp; 3 &amp; -0.7253154222 &amp; 0.2213356352 \\
-1.732050808 &amp; 3 &amp; 0.2476578246 &amp; -0.01533359952 \\
-3.464101615 &amp; 0 &amp; 0.9729732467 &amp; -0.2366692347
\end{array}\right)$$

Swapping the first row with the last:

$$\left(\begin{array}{ccc|c}
-3.464101615 &amp; 0 &amp; 0.9729732467 &amp; -0.2366692347 \\
-1.732050808 &amp; 3 &amp; 0.2476578246 &amp; -0.01533359952 \\
1.732050808 &amp; 3 &amp; -0.7253154222 &amp; 0.2213356352
\end{array}\right)$$

$\frac{1}{-3.464101615}R_1 \to R_1$:

$$\left(\begin{array}{ccc|c}
1 &amp; 0 &amp; -0.280873183 &amp; 0.06832052319 \\
-1.732050808 &amp; 3 &amp; 0.2476578246 &amp; -0.01533359952 \\
1.732050808 &amp; 3 &amp; -0.7253154222 &amp; 0.2213356352
\end{array}\right)$$

$1.732050808R_1 + R_2 \to R_2$
$-1.732050808R_1 + R_3 \to R_3$:

$$\left(\begin{array}{ccc|c}
1 &amp; 0 &amp; -0.280873183 &amp; 0.06832052319 \\
0 &amp; 3 &amp; -0.2388287988 &amp; 0.1030010178 \\
0 &amp; 3 &amp; -0.2388287988 &amp; 0.1030010178
\end{array}\right)$$

$\frac{1}{3} R_2 \to R_2$:

$$\left(\begin{array}{ccc|c}
1 &amp; 0 &amp; -0.280873183 &amp; 0.06832052319 \\
0 &amp; 1 &amp; -0.07960959961 &amp; 0.03433367262 \\
0 &amp; 3 &amp; -0.2388287988 &amp; 0.1030010178
\end{array}\right)$$

$-3R_2 + R_3 \to R_3$ (Final problematic matrix):

$$\left(\begin{array}{ccc|c}
1 &amp; 0 &amp; -0.280873183 &amp; 0.06832052319 \\
0 &amp; 1 &amp; -0.07960959961 &amp; 0.03433367262 \\
0 &amp; 0 &amp; 0 &amp; 0
\end{array}\right)$$

I derived this system of equations from an answer I recieved to a different question I asked about deriving the equation of a circle, $c$ (it's x, y, and r components) if you knew the equations of 3 other circles tangential to it, where one of the circles had radius = 0. Here is a link to the original question.

Someone posted an answer that alluded to the fact that I could derive a system of linear equations that would give me the solution I was after:

$$2(x_1-x_2)a+2(y_1-y_2)b+2(r_1-r_2)r=(x_1^2-x_2^2)+(y_1^2-y_2^2)-(r_1^2-r_2^2)$$
$$2(x_1-x_3)a+2(y_1-y_3)b+2r_1r=(x_1^2-x_3^2)+(y_1^2-y_3^2)-r_1^2$$
$$2(x_2-x_3)a+2(y_2-y_3)b+2r_2r=(x_2^2-x_3^2)+(y_2^2-y_3^2)-r_2^2$$

And from the graph I proposed in my question (shown below), it looked like there was only one solution, but I'm guessing my presumption was wrong. The green circle is the circle $c$ I'm trying to deduce:



So I'm wondering if it's possible to proceed with the Guassian Elimination despite the row of zeroes somehow?

Thank you!
Your system now reads

$$\begin{cases}x+az=b,\\y+cz=d.\end{cases}$$

Move $z$ to the RHS and you get a parametric solution

$$\begin{cases}x=b-az,\\y=d-cz.\end{cases}$$
You have obtained that one of your three equations is a linear combination of the other two, so it doesn’t give you any other information. So you have two equations in  three variables, so you have infinitely many solutions.

Maybe you didn’t use in the right way the Gaussian elimination: are you sure you have only linear variables? Can you show your system of equations (not matricial form)?