题目内容

14.已知矩阵A=$[\begin{array}{l}{1}&{-2}\\{-2}&{-1}\end{array}]$,B=$[\begin{array}{l}{5}\\{-15}\end{array}]$满足AX=B,求矩阵X.

分析 由AX=B,得$[\begin{array}{l}{1}&{-2}\\{-2}&{-1}\end{array}]$$[\begin{array}{l}{a}\\{b}\end{array}]$=$[\begin{array}{l}{5}\\{-15}\end{array}]$,求解即可.

解答 解:设x=$[\begin{array}{l}{a}\\{b}\end{array}]$,由$[\begin{array}{l}{1}&{-2}\\{-2}&{-1}\end{array}]$$[\begin{array}{l}{a}\\{b}\end{array}]$=$[\begin{array}{l}{5}\\{-15}\end{array}]$
得$\left\{\begin{array}{l}{a-2b=5}\\{-2a-b=-15}\end{array}\right.$解得$\left\{\begin{array}{l}{a=7}\\{b=1}\end{array}\right.$
此时x=$[\begin{array}{l}{7}\\{1}\end{array}]$

点评 本题主要考查了矩阵的应用,属于基础题型.

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