Matrix Median Mystery

using the laws of cosine: $\large m_1^2= \left(\dfrac{a}{2}\right)^2+b^2-2\dfrac{a}{2}b \cos(C) =\dfrac{a^2}{4}+b^2-ab \left(\dfrac{a^2+b^2-c^2}{2ab}\right) = \dfrac{1}{4} \left(-a^2+2b^2+2c^2\right)\\ \to \left[m_1^2\right]=\dfrac{1}{4} \begin{bmatrix} -1 && 2&&2\end{bmatrix} \begin{bmatrix} a^2 \\ b^2\\c^2\end{bmatrix}\to\begin{bmatrix} m_1^2 \\ m_2^2\\m_3^2\end{bmatrix}=\dfrac{1}{4} \begin{bmatrix} -1 && 2&&2\\2 && -1&&2\\2 && 2&&-1\end{bmatrix}\begin{bmatrix} a^2 \\ b^2\\c^2\end{bmatrix}$ where the symmetry of $(m_1,m_2,m_3)$ to $(a,b,c)$ was used to figure out the entire matrix from just one row, we inver this matrix to get $\large \begin{bmatrix} a^2 \\ b^2\\c^2\end{bmatrix} = \dfrac{4}{9}\begin{bmatrix} -1 && 2&&2\\2 && -1&&2\\2 && 2&&-1\end{bmatrix}\begin{bmatrix} m_1^2 \\ m_2^2\\m_3^2\end{bmatrix}$

Of course, given the existence of $M$ , we can apply the result to an equilateral triangle. All three sides have length $u$ , and all three medians have length $\tfrac12u\sqrt{3}$ , and then $3u^2 \; = \; a^2 + b^2 + c^2 \; = \; (1\;\;1\;\;1) M \left(\begin{array}{c} m_a^2 \\ m_b^2 \\ m_c^2 \end{array}\right) \; = \; (1\;\;1\;\;1) M \left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right) \tfrac34u^2 \; = \; X \times \tfrac34u^2$ where $X$ is the sum of the entries in $M$ , so that $X = \boxed{4}$ .

Note that M*[1, 1, 1] will equal to a 3x1 matrix, [a^2, b^2, c^2], in which each element in row i equals the sum of the entries in row i of M. So if we find a^2, b^2, and c^2 for which triangle ABC has medians of length 1, then a^2+b^2+c^2 is our answer. Since all of the medians are equal, you can safely assume that ABC is an equilateral triangle. Some basic geometry gives a = b = c = 2/sqrt(3) and the answer would be 4/3+4/3+4/3=4