例题 1.1

例题 1.1.

解方程组 $\{\begin{array}{ll}3x-2y+z& =15\\ 5y-2z& =9\\ 2x+y-z& =1\end{array}$

\begin{proof} 因为系数行列式为 $D=\left|\begin{array}{rrr}3& -2& 1\\ 0& 5& -2\\ 2& 1& -1\end{array}\right|=-11\ne 0,$ 所以方程组有解. 再由 $D_1=\left|\begin{array}{rrr}15& -2& 1\\ 9& 5& -2\\ 1& 1& -1\end{array}\right|=-55$

$D_2=\left|\begin{array}{rrr}3& 15& 1\\ 0& 9& -2\\ 2& 1& -1\end{array}\right|=-99$

$D_3=\left|\begin{array}{rrr}3& -2& 15\\ 0& 5& 9\\ 2& 1& 1\end{array}\right|=-198$

可得 $x=\frac{-55}{-11}=5,y=\frac{-99}{-11}=9,z=\frac{-198}{-11}=18.$ \end{proof}