pure,past paper,system of linear eqt

2008-02-17 7:54 pm
http://www2.hkedcity.net/sch_files/a/kst/kst-06185/public_html/sys.jpg

我想問解題方向.

once i look at this question, i just have no idea to do it, since we always treat it as a fact,rarely to prove it, so what can i do when i encounter this kind of question?

use this question as an example and explain the way of thinking plz.

回答 (1)

2008-02-17 9:29 pm
✔ 最佳答案
This question concerns the properties of the solutions of system of linear equations. To do this kinds of question, we have to be familar with the basic properties of system of linear equations. Once you get yourself familar with these properties, you would not find such questions difficult.

The solution (with some explanations) for this question is as follows
(i) The constant thems of system (I) are all zero. (Such a system is called a homogeneous system.) Such a system always has a solution, because (0,0,0) is obviously a solution.
The key point to do this question is to observe that system (II) can be obtained from system (I) by putting z=1.
If the system has a unique solution, then this solution must be (0,0,0). Now if system (II) has a solution, say, (x,y)=(u,v), then (x,y,z)=(u,v,1) will be a solution of system (I). However, the solution (u,v,1) is distinct from (0,0,0), (because the z-coordinate is different), which is impossible because it is given that the solution is unique. Hence, system (II) cannot have a solution.

(ii) If (u,v) is a solution of (II), then, (u,v,1) is a solution of (I). (We have just used this fact in part (i).) Now, multiplying the equations in system (I) by a constant t will not affect the R.H.S, since the R.H.S. is all 0. But then all terms in the L.H.S. is multiplied by t, and hence the solution (u,v,1) becomes the solution (ut, vt, t). This proves the “only if” part.
Conversely, if (ut, vt, t) is a solution of system (I), put t = 1 (in order to make the z-coordinate to be 1), we get a solution (u,v,1) of system (I). Hence (u,v) is a solution of system (II).

(iii) It is given that (I) has non-trivial solutions. Suppose (u,v,w) is a non-trivial solution of (I). Notice that we can get a solution of (II) if we can make the z-coordinate to be 1. We can do this if w≠0, because in that case, we can divide the solution by w, and get a solution (u/w, v/w, 1). But it is given that (II) has no solution, that means we cannot do this. Hence, we must have w=0.
Therefore, what we can say about the solution of (I) is that its z-coordinate must be 0, i.e., it must be of the form (u,v,0).

以上所用的方法都不是很深的東西,不過你一定要對system of linear equation 熟識而且靈活運用,才能解答得到。


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