75th Romanian Mathematical Olympiad

a) For given $i,j\in \{1,2,\dots ,n\}$, by computing the expansion along the row $i$, we obtain $\det (A+X_{ij})=\det (A)+{\delta }_{ij}$, where ${\delta }_{ij}$ is the $(i,j)$-cofactor of the matrix $A$. Since $A$ has the property $(\mathcal{P})$, it results that ${\delta }_{ij}={\delta }_{ji}$, for any $i,j\in \{1,2,\dots ,n\}$. Hence the adjugate matrix $A^{\ast }$ is symmetric. Let us denote $d=\det (A)\ne 0$. We have $A{A}^{\ast }=d{I}_n=(d{I}_n{)}^T=(A^{\ast }A{)}^T=A^T(A^{\ast }{)}^T=A^T{A}^{\ast }$. Since $d\ne 0$ and $d\det (A^{\ast })=\det (A{A}^{\ast })=\det (d{I}_n)=d^n$, we deduce that $A^{\ast }$ is invertible. Therefore, from the relation $A{A}^{\ast }=A^T{A}^{\ast }$, we obtain $A=A^T$.

b) Consider the matrix $A=(a_{ij}{)}_{1\le i,j\le n}\in {\mathcal{M}}_n(\mathbb{C})$ with $a_{11}=a_{12}=1$ and $a_{ij}=0$ elsewhere. Since $n\ge 3$, the matrix $A+X_{ij}$ has at least one null row, so $\det (A+X_{ij})=0$, for any $i,j\in \{1,2,\dots ,n\}$. Thus, $A$ has the property $(\mathcal{P})$, but $A\ne A^T$.