题号:6629    题型:填空题    来源:B站刘老师开讲《线性代数B》第七套期末模拟考试
$
\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|
$
0 人点赞 纠错 ​ ​ 12 次查看 ​ 我来讲解
答案:
答案:
-12

解析:

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

$\to$

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

$\to$

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

$$
\stackrel{r_1+(-3)}{=} 4 \times(-3)\left|\begin{array}{lll}
a_{11} & a_{21} & a_{31} \\
a_{12} & a_{22} & a_{32} \\
a_{13} & a_{23} & a_{33}
\end{array}\right|=-12\left|\boldsymbol{A}^{\mathrm{T}}\right|=-12|\boldsymbol{A}|=-12 \text {. }
$$
①点击 收藏 此题, 扫码注册关注公众号接收信息推送(一月四份试卷,中1+高2+研1)
② 程序开发、服务器资源都需要大量的钱,如果你感觉本站好或者受到到帮助,欢迎赞助本站,赞助方式:微信/支付宝转账到 18155261033

关闭


试题打分
①此题难易度如何

②此题推荐度如何

确定