and $BA = \left[ {\begin{array}{*{20}{c}}0&{ - i}\\i&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&i\end{array}} \right] = - AB$
$\therefore AB + BA = O$
Hence, ${(A + B)^2} = {A^2} + {B^2}$.
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$A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ ,
$n \in N$, then
$\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},$ where $\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}$
Then for the function $\mathrm{F}$, the point $\mathrm{x}=1$ is