Question

14．已知矩阵A=$[\begin{array}{l}{3}&{0}\\{2}&{1}\end{array}]$的逆矩阵A^-1=$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$，则行列式$|\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}|$的值为$\frac{1}{3}$．

Answer 1

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

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

点评 本题考查逆变换与逆矩阵，考查行列式的计算，考查计算能力，属于基础题．