# 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):

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

Swapping the first row with the last:

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

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

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

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

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

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

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

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

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

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!

• If you have a row of 0s, assuming your calculations were all correct, your Gaussian elimination is finished. The row of 0s indicates infinitely many solutions. – 79037662 Sep 24 '19 at 16:08
• I feel like linear algebra (in particular Gaussian elimination) is the wrong tool for this job. – 79037662 Sep 24 '19 at 16:11
• @79037662 Updated my question - the problem is to find the circle who is tangential to the other two circles and intersects the black point in the bottom right. These invariants provide me with a set of simultaneous equations which I'm trying to solve with Gaussian Elimination (with a computer). Do you have a different alternative you could suggest? I'm open to the idea! – jonny Sep 24 '19 at 16:16
• The reason I say this feels like the wrong tool is because of those nasty coefficients; are they (supposed to be) irrational numbers? If so, rounding them as you have done could lead to serious problems. Though I haven't studied this problem much, I expect there is a purely geometric way to find the answer; maybe that is worthy of its own question on MSE. – 79037662 Sep 24 '19 at 16:19
• The matrix is not singular. The singular values are $4.419739109997566e+00$, $4.242640687137096e+00$, and $1.002427530013128e-10$. The 2-norm condition number of your matrix is $4.409036042675010e+10$. The reason your Gaussian elimination is failing is because you are not using enough significant figures in your calculation. Your teacher has designed a good problem. Solve the problem using 16 significant figures (IEEE double precision) and you will be fine. – Carl Christian Sep 24 '19 at 16:21

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

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

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

• Note, the matrix is not singular, but so ill-conditioned that the suggested number of significant figures is insufficient. – Carl Christian Sep 24 '19 at 16:22
• Does this suggest that the system has inifinitely many solutions? I guess that's what I'm having trouble with. Looking at my graph, I'd think there was only one solution that satisifed what I stated... could that assumption be wrong? – jonny Sep 24 '19 at 16:24
• If your problem is to find a circle tangent to three given circles (one of which has zero radius), then there is a finite number of solutions (four if I am right) and the equations cannot be linear. – Yves Daoust Sep 24 '19 at 16:27
• That's right @YvesDaoust, that's the problem I'm trying to solve. I had initially posted another question and in the answer given it appeared that solving a set of linear equations might get me my answer. Is there a better approach to the problem? – jonny Sep 24 '19 at 16:38
• @jonny: the answer that you accepted is wrong, it suggests a linear system which is degenerate. As I said the system cannot be linear. Keep the first two equations that I wrote and plug them in equation (1) to get a quadratic one. – Yves Daoust Sep 24 '19 at 19:17

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)?