题目
1.求下列矩阵的逆矩阵.-|||-(I) a b] ,且 -cbneq 0 ;-|||-C d-|||-(2) (_(2)... (a)_(n)neq 0) (未写出的元素都是0).
题目解答
答案
本题考查矩阵的逆矩阵，属于基础题。
(1)由矩阵的逆矩阵公式可得结果；
(2)用初等行变换求逆矩阵；
(3)用初等列变换求逆矩阵。
(1)由矩阵的逆矩阵公式可得$\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)\left(\begin{array}{ll}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right)$=$\left(\begin{array}{ll}1& 0\\ 0& 1\end{array}\right)$，
所以$\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$的逆矩阵为$\left(\begin{array}{ll}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right)$；
(2)$\left(\$beginarraylll1$& 0& 0\\ 0& 1& 1\\ 1& 0& 1\$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& 0\\ 0& 1& 1\\ 0& 0& 1\$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& 0\\ 0& 1& 0\\ 0& 0& 1\$endarray$\right)$，
所以$\left(\begin{array}{lll}1& 0& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right)$的逆矩阵为$\left(\begin{array}{lll}1& 0& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right)$；
(3)$\left(\$beginarrayllla$&${a}_{2}$& \dots \\${a}_{1}$&${a}_{2}$& \dots \\ \dots & \dots & \dots \\${a}_{1}$&${a}_{2}$& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$&${a}_{2}$& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0&${a}_{2}$& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$& 0& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$& 0& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& \dots \\ 0& 1& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)$，
所以$\left(\begin{array}{lll}a& {a}_{2}& \dots \\ {a}_{1}& {a}_{2}& \dots \\ \dots & \dots & \dots \\ {a}_{1}& {a}_{2}& \dots \end{array}\right)$的逆矩阵为$\left(\begin{array}{lll}\frac{1}{{a}_{1}}& \frac{-{a}_{2}}{{{a}_{1}}^{2}}& \dots \\ 0& \frac{1}{{a}_{2}}& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \end{array}\right)$。
(1)由矩阵的逆矩阵公式可得结果；
(2)用初等行变换求逆矩阵；
(3)用初等列变换求逆矩阵。
(1)由矩阵的逆矩阵公式可得$\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)\left(\begin{array}{ll}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right)$=$\left(\begin{array}{ll}1& 0\\ 0& 1\end{array}\right)$，
所以$\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$的逆矩阵为$\left(\begin{array}{ll}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right)$；
(2)$\left(\$beginarraylll1$& 0& 0\\ 0& 1& 1\\ 1& 0& 1\$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& 0\\ 0& 1& 1\\ 0& 0& 1\$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& 0\\ 0& 1& 0\\ 0& 0& 1\$endarray$\right)$，
所以$\left(\begin{array}{lll}1& 0& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right)$的逆矩阵为$\left(\begin{array}{lll}1& 0& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right)$；
(3)$\left(\$beginarrayllla$&${a}_{2}$& \dots \\${a}_{1}$&${a}_{2}$& \dots \\ \dots & \dots & \dots \\${a}_{1}$&${a}_{2}$& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$&${a}_{2}$& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0&${a}_{2}$& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$& 0& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarrayllla$& 0& \dots \\ 0&${a}_{2}$& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)\$xrightarro$,
\left(\$beginarraylll1$& 0& \dots \\ 0& 1& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \$endarray$\right)$，
所以$\left(\begin{array}{lll}a& {a}_{2}& \dots \\ {a}_{1}& {a}_{2}& \dots \\ \dots & \dots & \dots \\ {a}_{1}& {a}_{2}& \dots \end{array}\right)$的逆矩阵为$\left(\begin{array}{lll}\frac{1}{{a}_{1}}& \frac{-{a}_{2}}{{{a}_{1}}^{2}}& \dots \\ 0& \frac{1}{{a}_{2}}& \dots \\ \dots & \dots & \dots \\ 0& 0& \dots \end{array}\right)$。