求代数余子式之和通常可以转化为求某行列式的值

一、题目

已知：

$$
|A|=\left|\begin{array}{cccc}1 & -5 & 1 & 3 \\ 1 & 1 & 3 & 4 \\ 1 & 1 & 2 & 3 \\ 2 & 0 & 1 & 5\end{array}\right|
$$

则：

$$
A_{41}+A_{42}+A_{43}+A_{44}=?
$$

难度评级：

二、解析

$$
A_{41}+A_{42}+A_{43}+A_{44}=
$$

$$
\left|\begin{array}{cccc}1 & -5 & 1 & 3 \\ 1 & 1 & 3 & 4 \\ 1 & 1 & 2 & 3 \\ 1 & 1 & 1 & 1\end{array}\right|=
$$

$$
\left|\begin{array}{cccc}1 & -5 & 1 & 3 \\ 0 & 6 & 2 & 1 \\ 0 & 6 & 1 & 0 \\ 0 & 6 & 0 & -2\end{array}\right|=(-1)^{1+1} \times 1 \times\left|\begin{array}{ccc}6 & 2 & 1 \\ 6 & 1 & 0 \\ 6 & 0 & -2\end{array}\right|=
$$

$$
\left|\begin{array}{ccc}6 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -2 & -3\end{array}\right|=(-1)^{1+1} \times 6 \times\left|\begin{array}{cc}-1 & -1 \\ -2 & -3\end{array}\right|=
$$

$$
(-1)^{1+1} \times 6 \times(-1) \times\left|\begin{array}{cc}1 & 1 \\ -2 & -3\end{array}\right|=
$$

$$
(-1)^{1+1} \times 6 \times(-1) \times(-1) \times\left|\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right|=
$$

$$
6 \times(3-2)=6
$$

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