$\text { 设 }\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|=1 \text {, 则 }\left|\begin{array}{ccc} 4 a_{11} & 4 a_{21} & 4 a_{31} \\ 2 a_{11}-3 a_{12} & 2 a_{21}-3 a_{22} & 2 a_{31}-3 a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}\right|$

-12

解析：

$$\text { 解 设 } \boldsymbol{A}=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \text {, 则已知 }|\boldsymbol{A}|=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|=1 \text {. }$$
$\to$
$$\left|\begin{array}{ccc} 4 a_{11} & 4 a_{21} & 4 a_{31} \\ 2 a_{11}-3 a_{12} & 2 a_{21}-3 a_{22} & 2 a_{31}-3 a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}\right|$$

$\to$

$$4\left|\begin{array}{ccc} a_{11} & a_{21} & a_{31} \\ 2 a_{11}-3 a_{12} & 2 a_{21} \text { đ } 3 a_{22} & 2 a_{31}-3 a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}\right|$$

$\to$

$$4\left|\begin{array}{ccc} a_{11} & a_{21} & a_{31} \\ -3 a_{12} & -3 a_{22} & -3 a_{32} \\ a_{13} & a_{23} & a_{23} \end{array}\right|$$

$$\stackrel{r_1+(-3)}{=} 4 \times(-3)\left|\begin{array}{lll} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}\right|=-12\left|\boldsymbol{A}^{\mathrm{T}}\right|=-12|\boldsymbol{A}|=-12 \text {. }$$
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