Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to show that the following system of equations does not have a solution, but I do not know how to do this

$$w_1+w_2=\frac{1}{2}$$ $$w_1s_1+w_2s_2=\frac{1}{6}$$ $$w_1t_1+w_2t_2=\frac{1}{6}$$ $$w_1s_1t_1+w_2s_2t_2=\frac{1}{24}$$ $$w_1s_1^2+w_2s_2^2=\frac{1}{12}$$ $$w_1t_1^2+w_2t_2^2=\frac{1}{12}$$ that all $t1,t2,w1,w2,s1,s2$ are unknown.

please help me, Thanks.

share|improve this question
    
this is having linear-algebra and matrices. Observe that the variables you have are in pairs, so the best you can do is use $2 \times 2$ matrices. You can split your system into pairs and find various conditions not to or to have solutions. It is a system of equations so you only need one contradiction to prove it has no solution. –  user31280 May 9 '13 at 6:19
add comment

1 Answer

up vote 1 down vote accepted

First note that we cannot have $w_1 = 0$ or $w_2 = 0$.

Subtracting the second and third give us

$$w_1(s_1 - t_1) = -w_2(s_2 - t_2)$$

Now if $s_1 = t_1$ then $s_2 = t_2$, and we can verify there is no solution, by comparing the fourth and fifth.

Subtracting the fifth and sixth gives us

$$w_1(s_1^2 - t_1^2) = -w_2(s_2^2 - t_2^2)$$

and thus from the above, we must have that

$$s_1 + t_1 = s_2 + t_2 = x \;(\text{say})$$

Adding second and third gives us

$$(w_1 + w_2) x = \frac{1}{3}$$

and so $$x = \frac{2}{3}$$ using the first equation.

Adding the fifth, sixth and two times the fourth gives us

$$w_1 x^2 + w_2 x^2 = \frac{1}{4}$$

and so $$x = \frac{\pm 1}{\sqrt{2}}$$

a contradiction.

share|improve this answer
    
Since we didn't use any inequalities, you can allow the variables to be complex too. –  Aryabhata May 9 '13 at 6:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.