设 $\boldsymbol{A}=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)$ 且 $|\boldsymbol{A}|=3, \boldsymbol{B}=\left(\begin{array}{lll}a_{13} & a_{12}+2 a_{11} & a_{11} \\ a_{23} & a_{22}+2 a_{21} & a_{21} \\ a_{33} & a_{32}+2 a_{31} & a_{31}\end{array}\right)$, 则 $\boldsymbol{B} \cdot \boldsymbol{A}=$