Question:

Step 1: The median length formulas for a triangle $\triangle ABC$ are given by: $$m_1^2 = \frac{2b^2+2c^2-a^2}{4}$$ $$m_2^2 = \frac{2a^2+2c^2-b^2}{4}$$ $$m_3^2 = \frac{2a^2+2b^2-c^2}{4}$$ Step 2: These equations can be rearranged to express $4m_i^2$ in terms of $a^2, b^2, c^2$: $$4m_1^2 = -a^2 + 2b^2 + 2c^2$$ $$4m_2^2 = 2a^2 - b^2 + 2c^2$$ $$4m_3^2 = 2a^2 + 2b^2 - c^2$$ This system can be written in matrix form as: $$4\begin{bmatrix}m_1^2\\m_2^2\\m_3^2\end{bmatrix} = \begin{bmatrix}-1&2&2\\2&-1&2\\2&2&-1\end{bmatrix}\begin{bmatrix}a^2\\b^2\\c^2\end{bmatrix}$$ Let $A = \begin{bmatrix}-1&2&2\\2&-1&2\\2&2&-1\end{bmatrix}$. Then $\begin{bmatrix}m_1^2\\m_2^2\\m_3^2\end{bmatrix} = \frac{1}{4}A\begin{bmatrix}a^2\\b^2\\c^2\end{bmatrix}$. Let $N = \frac{1}{4}A$. Step 3: The problem states $\begin{bmatrix}a^2\\b^2\\c^2\end{bmatrix} = M\begin{bmatrix}m_1^2\\m_2^2\\m_3^2\end{bmatrix}$. Thus, $M = N^{-1}$. Step 4: The inverse matrix $N^{-1}$ is determined. Since $N = \frac{1}{4}A$, $M = 4A^{-1}$. First, compute the determinant of $A$: $$\det(A) = -1((-1)(-1) - (2)(2)) - 2((2)(-1) - (2)(2)) + 2((2)(2) - (-1)(2))$$ $$\det(A) = -1(1-4) - 2(-2-4) + 2(4+2)$$ $$\det(A) = -1(-3) - 2(-6) + 2(6)$$ $$\det(A) = 3 + 12 + 12 = 27$$ Step 5: Next, compute the cofactor matrix of $A$: $$C_{11} = \det\begin{bmatrix}-1&2\\2&-1\end{bmatrix} = 1-4 = -3$$ $$C_{12} = -\det\begin{bmatrix}2&2\\2&-1\end{bmatrix} = -(-2-4) = 6$$ $$C_{13} = \det\begin{bmatrix}2&-1\\2&2\end{bmatrix} = 4+2 = 6$$ $$C_{21} = -\det\begin{bmatrix}2&2\\2&-1\end{bmatrix} = -(-2-4) = 6$$ $$C_{22} = \det\begin{bmatrix}-1&2\\2&-1\end{bmatrix} = 1-4 = -3$$ $$C_{23} = -\det\begin{bmatrix}-1&2\\2&2\end{bmatrix} = -(-2-4) = 6$$ $$C_{31} = \det\begin{bmatrix}2&2\\-1&2\end{bmatrix} = 4+2 = 6$$ $$C_{32} = -\det\begin{bmatrix}-1&2\\2&2\end{bmatrix} = -(-2-4) = 6$$ $$C_{33} = \det\begin{bmatrix}-1&2\\2&-1\end{bmatrix} = 1-4 = -3$$ The cofactor matrix is $\text{cof}(A) = \begin{bmatrix}-3&6&6\\6&-3&6\\6&6&-3\end{bmatrix}$. Step 6: The inverse of $A$ is $A^{-1} = \frac{1}{\det(A)}(\text{cof}(A))^T$: $$A^{-1} = \frac{1}{27}\begin{bmatrix}-3&6&6\\6&-3&6\\6&6&-3\end{bmatrix} = \frac{1}{9}\begin{bmatrix}-1&2&2\\2&-1&2\\2&2&-1\end{bmatrix}$$ Step 7: Thus, the matrix $M$ is: $$M = 4A^{-1} = \frac{4}{9}\begin{bmatrix}-1&2&2\\2&-1&2\\2&2&-1\end{bmatrix}$$ Step 8: The trace of $M$ is the sum of its diagonal elements: $$\text{tr}(M) = \frac{4}{9}(-1 + (-1) + (-1)) = \frac{4}{9}(-3) = -\frac{12}{9} = -\frac{4}{3}$$ Step 9: The matrix $M$ is symmetric. This is evident because $A$ is a symmetric matrix ($A=A^T$), and $M$ is a scalar multiple of $A$. Step 10: The determinant of $M$ is: $$\det(M) = \det\left(\frac{4}{9}A\right) = \left(\frac{4}{9}\right)^3\det(A) = \frac{64}{729} \cdot 27 = \frac{64 \cdot 27}{729} = \frac{64}{27}$$ Step 11: The sum of all elements of $M$ is: $$\text{sum}(M) = \frac{4}{9}(-1+2+2+2-1+2+2+2-1) = \frac{4}{9}(9) = 4$$