Question

14．已知矩陣A=$[\begin{array}{l}{3}&{0}\\{2}&{1}\end{array}]$的逆矩陣A^-1=$[\begin{array}{l}{a}&\\{c}&21gdcfb\end{array}]$，則行列式$|\begin{array}{l}{a}&\\{c}&1sljatb\end{array}|$的值為$\frac{1}{3}$．

Answer 1

分析 由A•A^-1═$[\begin{array}{l}{3}&{0}\\{2}&{1}\end{array}]$•$[\begin{array}{l}{a}&\\{c}&uzw6672\end{array}]$=E，列方程組求得逆矩陣A^-1，即可求得行列式$|\begin{array}{l}{a}&\\{c}&nttzzza\end{array}|$的值．

解答 解：由A•A^-1═$[\begin{array}{l}{3}&{0}\\{2}&{1}\end{array}]$•$[\begin{array}{l}{a}&\\{c}&psriimf\end{array}]$=E，
即：$\left\{\begin{array}{l}{3a=1}\\{3b=0}\\{2a+c=0}\\{2b+d=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{1}{3}}\\{b=0}\\{c=-\frac{2}{3}}\\{d=1}\end{array}\right.$，
$|\begin{array}{l}{a}&\\{c}&a22ytb7\end{array}|$=$|\begin{array}{l}{\frac{1}{3}}&{0}\\{-\frac{2}{3}}&{1}\end{array}|$=$\frac{1}{3}$，
故答案為：$\frac{1}{3}$．

點評 本題考查逆變換與逆矩陣，考查行列式的計算，考查計算能力，屬于基礎題．