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For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.

The expression assumes $Y$ is invertible. By the cyclic property of the trace,For a real or complex antisymmetric matrix, $X^T=-X$ forces every diagonal entry to satisfy $x_{ii}=-x_{ii}$, hence $x_{ii}=0$. ThereforeThis also demonstrates invariance of the matrix trace under matrix similarity. Antisymmetry of $Z$ itself is not needed.

Let $H(u)={\sin }^{2}u+\ln u$. The chain rule gives $H^{\prime }(u)=\sin 2u+1/u$, so the two given summands combine into its logarithmic derivative, $f(u)=H^{\prime }(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi )=\ln \pi$,This satisfies $F(\pi )=0$, and differentiation recovers both terms of the integrand.

The fundamental theorem of calculus, using continuity of $f$ near $x$, givesThe lower endpoint $a$ is constant, so it contributes no additional term.

Of the ten outcomes in $F$, all except $(1,1,1)$ have exactly one repeated pair. Hence $P(T\mid F)=9/10$. Applying Bayes' theorem in the requested direction,Equivalently there are nine favorable outcomes among the ninety outcomes in $T$. This direct count confirms the conditional probability calculation.