题目内容

2.若$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$是方程组$\left\{\begin{array}{l}{kx-my=1}\\{mx+ky=8}\end{array}\right.$的解,则k,m的值为(  )
A.$\left\{\begin{array}{l}{k=2}\\{m=3}\end{array}\right.$B.$\left\{\begin{array}{l}{k=2}\\{m=-3}\end{array}\right.$C.$\left\{\begin{array}{l}{k=3}\\{m=2}\end{array}\right.$D.$\left\{\begin{array}{l}{k=-3}\\{m=-2}\end{array}\right.$

分析 把方程组的解代入方程组,再解关于m、k的方程组即可.

解答 解:把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{kx-my=1}\\{mx+ky=8}\end{array}\right.$得:$\left\{\begin{array}{l}{2k-m=1①}\\{2m+k=8②}\end{array}\right.$,
①×2+②得:5k=10,
解得:k=2,
把k=2代入①得:4-m=1,
解得:m=3,
故选A.

点评 本题考查了二元一次方程组的解,解二元一次方程组的应用,能得出关于m、k的方程组是解此题的关键.

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