# 2023年考研数二第16题解析：非齐次线性方程组、矩阵的子式、行列式的按行按列展开

## 二、解析

$$r(A) = r(A, b)$$

$$r(A) = r(A, b) \leqslant 3$$

$$\left[\begin{array}{llll}\textcolor{orangered}{a} & \textcolor{orangered}{0} & \textcolor{orangered}{1} & 1 \\ \textcolor{orangered}{1} & \textcolor{orangered}{a} & \textcolor{orangered}{1} & 0 \\ \textcolor{orangered}{1} & \textcolor{orangered}{2} & \textcolor{orangered}{a} & 0 \\ a & b & 0 & 2\end{array}\right] \Rightarrow \left[\begin{array}{lll}\textcolor{orangered}{a} & \textcolor{orangered}{0} & \textcolor{orangered}{1} \\ \textcolor{orangered}{1} & \textcolor{orangered}{a} & \textcolor{orangered}{1} \\ \textcolor{orangered}{1} & \textcolor{orangered}{2} & \textcolor{orangered}{a} \end{array}\right]$$

\begin{aligned} |A, b|= \\ & \left|\begin{array}{llll}a & 0 & 1 & \textcolor{springgreen}{1} \\ 1 & a & 1 & 0 \\ 1 & 2 & a & 0 \\ a & b & 0 & \textcolor{springgreen}{2} \end{array}\right|= \\ \\ & \textcolor{springgreen}{1} \cdot(-1)^{1+4}\left|\begin{array}{ccc}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right| + \textcolor{springgreen}{2} \cdot(-1)^{4+4}\left|\begin{array}{ccc}a & 0 & 1 \\ 1 & a & 1 \\ 1 & 2 & a\end{array}\right|= \\ \\ & -\left|\begin{array}{ccc}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|+2 \cdot 4=0 \end{aligned}

$$\left|\begin{array}{lll}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|=8$$