lead113

i forgot the approach ,  thnx ar

[ 本帖最後由 lead113 於 2014-4-17 03:43 PM 編輯 ]

LAnews

It asks for at least one solution, so check the determinant first
your determinant is now (s+2t)^2

if s=/= -2t, then you have unique solution, i.e fulfill at least one solution

if s = -2t, then your system will be

2tx + 4ty = 7
tx + 2ty = 3

and

tx + 2ty = 7/2
tx + 2ty = 3

giving you a contradicting result.

So you should choose any value as of s and t, not including s=-2t will be alright.
Hope i can help you

[ 本帖最後由 LAnews 於 2014-4-17 05:28 PM 編輯 ]

lead113

引用:

原帖由 LAnews 於 2014-4-17 04:49 PM 發表
So you should choose any value as of s and t, not including s=-2t will be alright.

thanks, but for fixed value, isnt it means we have to find a particular value?

桃子

引用:

原帖由 lead113 於 2014-4-17 06:09 PM 發表

thanks, but for fixed value, isnt it means we have to find a particular value?

It means s,t are constant.

lead113

that means we can take any value as a constant value? seems a bit strange

[ 本帖最後由 lead113 於 2014-4-24 09:53 PM 編輯 ]

steveabc

係呢度係 姐係唔只一個答案 只要s=/= -2t 無論s同t係咩 你都會出到unique solution, 因為det=/=0