IMC 2014 · Day 1 · P1

Let the matrix be $$M = \begin{bmatrix} x & z \\ z & y \end{bmatrix}.$$ The two conditions give us $x + y = a$ and $xy - z^2 = b$. Since this is symmetric in $x$ and $y$, the matrix can only be unique if $x = y$. Hence $2x = a$ and $x^2 - z^2 = b$. Moreover, if $(x, y, z)$ solves the system of equations, so does $(x, y, -z)$. So $M$ can only be unique if $z = 0$. This means that $2x = a$ and $x^2 = b$, so $a^2 = 4b$.

If this is the case, then $M$ is indeed unique: if $x + y = a$ and $xy - z^2 = b$, then $$(x - y)^2 + 4z^2 = (x + y)^2 + 4z^2 - 4xy = a^2 - 4b = 0,$$ so we must have $x = y$ and $z = 0$, meaning that $$M = \begin{bmatrix} a/2 & 0 \\ 0 & a/2 \end{bmatrix}$$ is the only solution.

Note that $\operatorname{trace}(M) = a$ and $\det(M) = b$ if and only if the two eigenvalues $\lambda_1$ and $\lambda_2$ of $M$ are solutions of $x^2 - ax + b = 0$. If $\lambda_1 \ne \lambda_2$, then $$M_1 = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \quad \text{and} \quad M_2 = \begin{bmatrix} \lambda_2 & 0 \\ 0 & \lambda_1 \end{bmatrix}$$ are two distinct solutions, contradicting uniqueness. Thus $\lambda_1 = \lambda_2 = \lambda = a/2$, which implies $a^2 = 4b$ once again. In this case, we use the fact that $M$ has to be diagonalisable as it is assumed to be symmetric. Thus there exists a matrix $T$ such that $$M = T^{-1} \cdot \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \cdot T,$$ however this reduces to $M = \lambda (T^{-1} \cdot I \cdot T) = \lambda I$, which shows again that $M$ is unique.