 Solution

Multiplicativity of the determinant gives $⟨A,B⟩=\det A\det B$. Thus C1 holds. For $n\ge 2$, C2 fails: choosing $A=B=I$ and $\lambda =2$ gives $\det (2I)=2^n\ne 2$. The same choice $A=B=C=I$ disproves C3, since $\det (B+C)=2^n\ne 2$. The nonnegativity portion of C4 holds, because $\det (A^2)=(\det A{)}^2\ge 0$, but its positive-definiteness clause fails: a nonzero singular matrix that is symmetric, such as $\mathrm{diag}(1,0,\dots ,0)$, has zero self-pairing.

Therefore this is not an inner product for $n\ge 2$. For the exceptional case $n=1$, the determinant is the scalar entry and the pairing is ordinary multiplication of real numbers, so all four conditions hold. The determinant pairing is not an inner product in higher dimension because it measures volume scaling rather than a bilinear length geometry.