Question

9．已知矩阵$A=[{\begin{array}{l}2&1\\ 3&2\end{array}}]$，列向量$X=[{\begin{array}{l}x\\ y\end{array}}]，B=[{\begin{array}{l}4\\ 7\end{array}}]$，若AX=B，直接写出A^-1，并求出X．

Answer 1

分析 法一：由矩阵$A=[{\begin{array}{l}2&1\\ 3&2\end{array}}]$，得A^-1=$[\begin{array}{l}{2}&{-1}\\{-3}&{2}\end{array}]$，由AX=B，得X=A^-1B，由此能求出X．
法二：由矩阵$A=[{\begin{array}{l}2&1\\ 3&2\end{array}}]$，得A^-1=$[\begin{array}{l}{2}&{-1}\\{-3}&{2}\end{array}]$，由AX=B，列出方程组，求出x，y，由此能求出X．

解答 解法一：∵矩阵$A=[{\begin{array}{l}2&1\\ 3&2\end{array}}]$，∴A^-1=$[\begin{array}{l}{2}&{-1}\\{-3}&{2}\end{array}]$，
∵AX=B，
∴X=A^-1B=$[\begin{array}{l}{2}&{-1}\\{-3}&{2}\end{array}]$$[\begin{array}{l}{4}\\{7}\end{array}]$=$[\begin{array}{l}{1}\\{2}\end{array}]$．
解法二：∵矩阵$A=[{\begin{array}{l}2&1\\ 3&2\end{array}}]$，∴A^-1=$[\begin{array}{l}{2}&{-1}\\{-3}&{2}\end{array}]$，
∵AX=B，
∴$[\begin{array}{l}{2}&{1}\\{3}&{2}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{4}\\{7}\end{array}]$，
∴$\left\{\begin{array}{l}{2x+y=4}\\{3x+2y=7}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，
∴X=$[\begin{array}{l}{1}\\{2}\end{array}]$．

点评 本考查逆矩阵的求法，考查矩阵方程的解法，是基础题，解题时要认真审题，注意矩阵性质的合理运用．