已知线性方程组 $\left\{\begin{array}{c}a x_1+x_3=1 \\ x_1+a x_2+x_3=0 \\ x_1+2 x_2+a x_3=0 \\ a x_1+b x_2=2\end{array}\right.$ 有解, 其中 $\mathrm{a}, \mathrm{b}$ 为常数, 若 $\left|\begin{array}{lll}a & 0 & 1 \\ 1 & a & 1 \\ 1 & 2 & a\end{array}\right|=4$, 则 $\left|\begin{array}{lll}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|=$
【答案】 8

【解析】 由已知 $r(A)=r(A, b) \leq 3 < 4$, 故 $|A, b|=0$
即 $|A, b|=\left|\begin{array}{llll}a & 0 & 1 & 1 \\ 1 & a & 1 & 0 \\ 1 & 2 & a & 0 \\ a & b & 0 & 2\end{array}\right|=1 \cdot(-1)^{1+4}\left|\begin{array}{lll}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|+2 \cdot(-1)^{4+4}\left|\begin{array}{lll}a & 0 & 1 \\ 1 & a & 1 \\ 1 & 2 & a\end{array}\right|=-\left|\begin{array}{lll}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|+2 \cdot 4=0$, 故 $\left|\begin{array}{lll}1 & a & 1 \\ 1 & 2 & a \\ a & b & 0\end{array}\right|=8$.
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