What is the sum of the coefficients of the equation?

See Steiner ellipse, minimal area through three points and Affine transformations . What works in two dimensions works in one dimensional (but then the problem is trivial) and in higher dimensions than two-dimensional. The methodology remains the same.

The equation with rounded coefficients is $1. \mathbf{x}_1^2-0.777254 \mathbf{x}_2 \mathbf{x}_1-1.18091 \mathbf{x}_3 \mathbf{x}_1-2.94991 \mathbf{x}_4 \mathbf{x}_1+5.81556 \mathbf{x}_1+0.255156 \mathbf{x}_2^2+0.545669 \mathbf{x}_3^2+ \\ 2.3023 \mathbf{x}_4^2-4.10312 \mathbf{x}_2+0.64231 \mathbf{x}_2 \mathbf{x}_3-5.93282 \mathbf{x}_3+1.30053 \mathbf{x}_2 \mathbf{x}_4+1.89688 \mathbf{x}_3 \mathbf{x}_4-10.2905 \mathbf{x}_4+17.8745 \\ =0$

The ends of the ellipsoid's axes with rounded coordinates: $\begin{array}{|l|l|} \hline \{2.42044,3.03429,5.54365,0.418117\} & \{5.17956,5.36571,1.65635,3.58188\} \\ \{5.13547,5.89936,3.37647,2.12303\} & \{2.46453,2.50064,3.82353,1.87697\} \\ \{5.27843,1.68465,3.20862,3.53262\} & \{2.32157,6.71535,3.99138,0.467384\} \\ \{0.772199,6.31337,2.26618,-0.175684\} & \{6.8278,2.08663,4.93382,4.17568\} \\ \hline \end{array}$

OK, OK OK, how did I get there:

The $\bm{n}$ array, the regular 5-cell vertex homogeneous coordinates in radial form and which locations are on a unit four-dimension sphere it is easy to prove that the unit regular 5-cell has the maximum volume within that sphere: $\left( \begin{array}{|c|c|c|c|c|} \hline 1 & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \hline 0 & \frac{\sqrt{15}}{4} & -\frac{\sqrt{\frac{5}{3}}}{4} & -\frac{\sqrt{\frac{5}{3}}}{4} & -\frac{\sqrt{\frac{5}{3}}}{4} \\ \hline 0 & 0 & \sqrt{\frac{5}{6}} & -\frac{\sqrt{\frac{5}{6}}}{2} & -\frac{\sqrt{\frac{5}{6}}}{2} \\ \hline 0 & 0 & 0 & -\frac{\sqrt{\frac{5}{2}}}{2} & \frac{\sqrt{\frac{5}{2}}}{2} \\ \hline \end{array} \right)$

From here on, I am providing only the numeric forms of the matrices as the radical forms take minutes to compute and take up pages to present.

The same matrix in numeric form: \left( \begin{array}{|l|l|l|l|l|} \hline 1. & -0.25 & -0.25 & -0.25 & -0.25 \ \hline 0. & 0.968246 & -0.322749 & -0.322749 & -0.322749 \ \hline 0. & 0. & 0.912871 & -0.456435 & -0.456435 \ \hline 0. & 0. & 0. & -0.790569 & 0.790569 \ 1. & 1. & 1. & 1. & 1. \ \hline \end{array} \right)]

The inverse of $\bm{n}$ in numeric form: $\left( \begin{array}{|l|l|l|l|l|} \hline 0.8 & 0 & 0 & 0 & 0.2 \\ \hline -0.2 & 0.774597 & 0 & 0 & 0.2 \\ \hline -0.2 & -0.258199 & 0.730297 & 0 & 0.2 \\ \hline -0.2 & -0.258199 & -0.365148 & -0.632456 & 0.2 \\ \hline -0.2 & -0.258199 & -0.365148 & 0.632456 & 0.2 \\ \hline \end{array} \right)$

The equation for a four-dimensional sphere: $\mathbf{x}_1^2+\mathbf{x}_2^2+\mathbf{x}_3^2+\mathbf{x}_4^2=1$ .

The $\bm{o}$ , which contains the five four-dimensional points in homogeneous coordinate form: $\left( \begin{array}{|l|l|l|l|l|} \hline 2. & 2. & 2. & 5. & 7. \\ \hline 7. & 3. & 5. & 3. & 2. \\ \hline 2. & 3. & 2. & 3. & 3. \\ \hline 1. & 1. & 1. & 1. & 5. \\ \hline 1. & 1. & 1. & 1. & 1. \\ \hline \end{array} \right)$

At this point, the determinant of $\bm{o}$ is checked to be non-zero so that it is invertible, i. e., the points are not contained in a subspace.

At this point both $\bm{o}$ and $\bm{n}$ are known and therefore $\bm{a}$ can be computed from the $\bm{a}.\bm{o}=\bm{n}$ by inverting $\bm{o}$ and right multiplying $\bm{n}$ giving $\bm{a}$ : \[\left( \begin{array}{|l|l|l|l|l|} \hline -0.707547 & 0.377358 & 0.212264 & 0.990566 & -1.64151 \\ \hline -0.645497 & 0. & 0.322749 & 0.645497 & 0. \\ \hline 0.766467 & -0.439211 & -0.732019 & -1.4382 & 4.44379 \\ \hline -0.074582 & -0.223746 & -0.37291 & 0.104415 & 2.35679 \\ \hline 0. & 0. & 0. & 0. & 1. \\ \hline

\end{array} \right)\]

The inverse of $\bm{a}$ can be computed by inverting $\bm{a}$ : $\left( \begin{array}{|l|l|l|l|l|} \hline -1.8 & -2.32379 & -2.19089 & 1.26491 & 3.8 \\ \hline 2.8 & -1.54919 & 1.09545 & -1.89737 & 4.2 \\ \hline -1.6 & 1.0328 & -0.730297 & -1.26491 & 3.6 \\ \hline -1. & -1.29099 & -1.82574 & 1.89737 & 2. \\ \hline 0. & 0. & 0. & 0. & 1. \\ \hline \end{array} \right)$

At this point, the Singular Value Decomposition of the affine matrix portion of the inverse of $\bm{a}$ is computed to get the scale factors of the ellipsoid's axes (the diagonal of the $\bm{\sigma}$ and the rotation to the directions of those axes (\bm{v}). By the way, this is where the eigenvalues and eigenvectors mentioned by Mark Henning come in to the solution: $\bm{u}=\left( \begin{array}{|l|l|l|l|} \hline 0.644939 & 0.418597 & -0.401755 & 0.497417 \\ \hline -0.540846 & 0.692496 & -0.42234 & -0.222631 \\ \hline 0.0969208 & -0.510956 & -0.787184 & -0.331469 \\ \hline 0.531174 & 0.290087 & 0.201405 & -0.770156 \\ \hline \end{array} \right)$ $\text{diagonal}(\bm{\sigma})=\{5.43777,3.44857,1.95091,0.414642\}$ $\bm{v}=\left( \begin{array}{|l|l|l|l|} \hline -0.618177 & 0.496715 & 0.306881 & -0.526266 \\ \hline -0.229225 & -0.854775 & 0.26391 & -0.383625 \\ \hline -0.560161 & -0.0913368 & 0.320217 & 0.758511 \\ \hline 0.501531 & 0.119552 & 0.856526 & 0.0231813 \\ \hline \end{array} \right)$

Yes, I checked the $\bm{a}$ and $\bm{a}^{-1}$ were, in fact, inverses of each other.

Using $\bm{a}$ , $\left\{\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3,\mathbf{x}_4\right\}$ were projected into the unit sphere environment and substituted into the Unit sphere equation ( $\mathbf{x}_1^2+\mathbf{x}_2^2+\mathbf{x}_3^2+\mathbf{x}_4^2=1$ ) and then the resultant equation was simplified. The resultant equation was given above and not repeated here.

Forward mapping the positive and negative unit vectors using the $\bm{v}$ matrix and using the inverse of the $\bm{a}$ back the original environment of the specified points gives the ends of the axes of the ellipsoid. These also were provided above and not repeated here